NCERT Solutions Class 10 Maths Chapter 8

Q.1

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B, AB}=\text{24 cm, BC}=\text{7 cm.}\\ \text{Determine :}\\ \text{(i) sin A, cos A}\\ \text{(ii) sin C, cos C}\end{array}$

Ans.

$\begin{array}{l}\text{Using Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ \text{AC\hspace{0.17em}}=\sqrt{{24}^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\text{cm}\end{array}$

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}\\ \cos \text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\ \text{(ii)}\\ \text{sin\hspace{0.17em}C}=\frac{\text{side opposite to angle C}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\ \cos \text{C}=\frac{\text{side adjacent to angle C}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}\end{array}$

Q.2

$\text{In the following figure},\mathrm{tanP}-\mathrm{cotR}.$

Ans.

$\begin{array}{l}\text{Using Pythagoras theorem in}\mathrm{\Delta }\text{PQR, we get}\\ \text{QR\hspace{0.17em}}=\sqrt{{13}^2-{12}^2}=\sqrt{169-144}=\sqrt{25}=5\text{cm}\end{array}$

$\begin{array}{l}\mathrm{tanP}=\frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\ \mathrm{cotR}=\frac{\text{side adjacent to angle R}}{\text{side opposite to angle R}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\ \\ \mathrm{tanP}-\mathrm{cotR}=\frac{5}{12}-\frac{5}{12}=0\end{array}$

Q.3

$\text{If sin A}=\frac{3}{4},\text{calculate}\cos \text{A and}\tan \text{A.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right-angled triangle right angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{sin A}=\frac{3}{4}\\ \text{or}\frac{\text{BC}}{\text{AC}}=\frac{3}{4}\\ \text{Let BC be 3k. Therefore, AC will be 4k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{or (4k)}}^2={\text{AB}}^2+{\left(3\mathrm{k}\right)}^2\\ \text{or}16{\mathrm{k}}^2-9{\mathrm{k}}^2={\text{AB}}^2\\ \text{or \hspace{0.17em}}7{\mathrm{k}}^2={\text{AB}}^2\\ \text{or \hspace{0.17em} AB}=\sqrt{7}\mathrm{k}\\ \cos \text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{7}\mathrm{k}}{4\mathrm{k}}=\frac{\sqrt{7}}{4}\\ \tan \text{A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{\text{BC}}{\text{AB}}=\frac{3\mathrm{k}}{\sqrt{7}\mathrm{k}}=\frac{3}{\sqrt{7}}\end{array}$

Q.4

$\text{Given 15 cot A}=\text{8, find sin A and sec A.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right-angled triangle right angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{15 cot A}=\text{8}\\ \text{or cot A}=\frac{8}{15}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{8}{15}\\ \text{Let AB be 8k. Therefore, BC will be 15k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{or AC}}^2={\left(\text{15k}\right)}^2+{\left(8\mathrm{k}\right)}^2\\ {\text{or AC}}^2=225{\mathrm{k}}^2+64{\mathrm{k}}^2\\ {\text{or \hspace{0.17em}AC}}^2=\text{289}{\mathrm{k}}^2\\ \text{or \hspace{0.17em}AC}=17\mathrm{k}\\ \sin \text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15\mathrm{k}}{17\mathrm{k}}=\frac{15}{17}\\ \sec \text{A}=\frac{\text{hypotenuse}}{\text{side adjacent to angle A}}=\frac{\text{AC}}{\text{AB}}=\frac{17\mathrm{k}}{8\mathrm{k}}=\frac{17}{8}\end{array}$

Q.5

$\text{Given sec}\mathrm{\theta }=\frac{13}{12}\text{, calculate all other trigonometric ratios.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{sec}\mathrm{\theta }=\frac{13}{12}\\ \text{or}\frac{\text{AC}}{\text{AB}}=\frac{13}{12}\\ \text{Let AC be 13k. Therefore, AB is 12k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{BC}}^2={\text{AC}}^2-{\text{AB}}^2\\ {\text{or BC}}^2={\left(\text{13k}\right)}^2-{\left(12\mathrm{k}\right)}^2\\ {\text{or BC}}^2=169{\mathrm{k}}^2-144{\mathrm{k}}^2\\ {\text{or \hspace{0.17em}BC}}^2=\text{25}{\mathrm{k}}^2\\ \text{or \hspace{0.17em}BC}=5\mathrm{k}\\ \mathrm{sin\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{5\mathrm{k}}{13\mathrm{k}}=\frac{5}{13}\\ \mathrm{cos\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12\mathrm{k}}{13\mathrm{k}}=\frac{12}{13}\\ \mathrm{tan\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{side adjacent to angle}\mathrm{\theta }}=\frac{\text{BC}}{\text{AB}}=\frac{5\mathrm{k}}{12\mathrm{k}}=\frac{5}{12}\\ \mathrm{cot\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{side opposite to angle}\mathrm{\theta }}=\frac{\text{AB}}{\text{BC}}=\frac{12\mathrm{k}}{5\mathrm{k}}=\frac{12}{5}\\ \mathrm{cosec}\mathrm{\theta }=\frac{\text{hypotenuse}}{\text{side opposite to angle}\mathrm{\theta }}=\frac{\text{AC}}{\text{BC}}=\frac{13\mathrm{k}}{5\mathrm{k}}=\frac{13}{5}\end{array}$

Q.6

$\begin{array}{l}\text{If}\angle \text{A and}\angle \text{B are acute angles such that cos A}=\text{cos B,}\\ \text{then show that}\angle \text{A}=\angle \text{B.}\end{array}$

Ans.

$\text{Let us consider a}\mathrm{\Delta }\text{\hspace{0.17em}ABC in which CD}\perp \text{AB.}$

$\begin{array}{l}\text{Given that,}\\ \text{cos A}=\text{cos B}\\ \text{or}\frac{\text{AD}}{\text{AC}}=\frac{\text{BD}}{\text{BC}}\\ \text{or}\frac{\text{AD}}{\text{BD}}=\frac{\text{AC}}{\text{BC}}\\ \text{Let}\frac{\text{AD}}{\text{BD}}=\frac{\text{AC}}{\text{BC}}=\mathrm{k}\\ \text{or AD}=\mathrm{k}\text{BD}...\text{(1)}\\ \text{and}\\ \text{AC}=\mathrm{k}\text{BC}...\text{(2)}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{\hspace{0.17em}CAD and}\mathrm{\Delta }\text{\hspace{0.17em}CBD, we get}\\ {\text{CD}}^2={\text{AC}}^2-{\text{AD}}^2...\text{(3)}\\ \text{and}\\ {\text{CD}}^2={\text{BC}}^2-{\text{BD}}^2...\text{(4)}\\ \text{From equations (3) and (4), we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^2-{\text{AD}}^2={\text{BC}}^2-{\text{BD}}^2\\ \text{(}\mathrm{k}{\text{BC)}}^2-{\left(\mathrm{k}\text{BD}\right)}^2={\text{BC}}^2-{\text{BD}}^2\\ \text{or}{\mathrm{k}}^2{\text{(BC}}^2-{\text{BD}}^2)={\text{BC}}^2-{\text{BD}}^2\\ \text{or}{\mathrm{k}}^2=1\\ \text{or \hspace{0.17em}}\mathrm{k}=1\\ \text{On putting this value of}\mathrm{k}\text{in equation (2), we get}\\ \text{AC}=\text{BC}\\ \text{i.e.,}\angle \text{A}=\angle \text{B}\end{array}$

Q.7

$\text{If cot}\mathrm{\theta }=\frac{\text{7}}{8},\text{evaluate : (i)}\frac{(1+\mathrm{sin\theta })(1-\mathrm{sin\theta })}{(1+\mathrm{cos\theta })(1-\mathrm{cos\theta })},{\text{(ii) cot}}^2\mathrm{\theta }$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{cot}\mathrm{\theta }=\frac{7}{8}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{7}{8}\\ \text{Let BC be 8k. Therefore, AB is 7k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{or AC}}^2={\left(\text{7k}\right)}^2+{\left(8\mathrm{k}\right)}^2\\ {\text{or AC}}^2=49{\mathrm{k}}^2+64{\mathrm{k}}^2\\ {\text{or \hspace{0.17em}AC}}^2=11\text{3}{\mathrm{k}}^2\\ \text{or \hspace{0.17em}AC}=\sqrt{113}\mathrm{k}\\ \mathrm{sin\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{8\mathrm{k}}{\sqrt{113}\mathrm{k}}=\frac{8}{\sqrt{113}}\\ \mathrm{cos\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{7\mathrm{k}}{\sqrt{113}\mathrm{k}}=\frac{7}{\sqrt{113}}\\ \text{(i)}\frac{(1+\mathrm{sin\theta })(1-\mathrm{sin\theta })}{(1+\mathrm{cos\theta })(1-\mathrm{cos\theta })}=\frac{1-{\sin }^2\mathrm{\theta }}{1-{\cos }^2\mathrm{\theta }}=\frac{1-{\left(\frac{8}{\sqrt{113}}\right)}^2}{1-{\left(\frac{7}{\sqrt{113}}\right)}^2}=\frac{113-64}{113-49}=\frac{49}{64}\\ {\text{(ii) cot}}^2\mathrm{\theta }={\left(\frac{7}{8}\right)}^2=\frac{49}{64}\end{array}$

Q.8

$\text{If 3 cot A}=\text{4, check whether}\frac{1-{\tan }^2\text{A}}{1+{\tan }^2\text{A}}={\cos }^2\text{A}-{\sin }^2\text{A or not.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{cot A}=\frac{4}{3}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{4}{3}\\ \text{Let BC be 3k. Therefore, AB = 4k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{or AC}}^2={\left(\text{4k}\right)}^2+{\left(3\mathrm{k}\right)}^2\\ {\text{or AC}}^2=16{\mathrm{k}}^2+9{\mathrm{k}}^2\\ {\text{or \hspace{0.17em}AC}}^2=25{\mathrm{k}}^2\\ \text{or \hspace{0.17em}AC}=5\mathrm{k}\\ \sin \text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{3\mathrm{k}}{5\mathrm{k}}=\frac{3}{5}\\ \cos \text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{4\mathrm{k}}{5\mathrm{k}}=\frac{4}{5}\\ \tan \text{A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{\text{BC}}{\text{AB}}=\frac{3\mathrm{k}}{4\mathrm{k}}=\frac{3}{4}\\ \text{Now,}\\ \frac{1-{\tan }^2\text{A}}{1+{\tan }^2\text{A}}=\frac{1-{\left(\frac{3}{4}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}=\frac{7}{25}\\ {\cos }^2\text{A}-{\sin }^2\text{A}={\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2=\frac{7}{25}\\ \therefore \frac{1-{\tan }^2\text{A}}{1+{\tan }^2\text{A}}={\cos }^2\text{A}-{\sin }^2\text{A}\end{array}$

Q.9

$\begin{array}{l}\text{In triangle ABC, right-angled at B, if tan A}=\frac{1}{\sqrt{3}},\text{find the value of:}\\ \text{(i) sin A cos C}+\text{cos A sin C}\\ \text{(ii) cos A cos C}-\text{sin A sin C}\end{array}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{tan A}=\frac{1}{\sqrt{3}}\\ \text{or}\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{3}}\\ \text{Let BC be}\mathrm{k}\text{. Then, AB =}\sqrt{3}\mathrm{k}\text{, where}\mathrm{k}\text{is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{or AC}}^2={\left(\sqrt{3}\text{k}\right)}^2+{\left(\mathrm{k}\right)}^2\\ {\text{or AC}}^2=3{\mathrm{k}}^2+{\mathrm{k}}^2\\ {\text{or \hspace{0.17em}AC}}^2=4{\mathrm{k}}^2\\ \text{or \hspace{0.17em}AC}=2\mathrm{k}\\ \sin \text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\mathrm{k}}{2\mathrm{k}}=\frac{1}{2}\\ \cos \text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\mathrm{k}}{2\mathrm{k}}=\frac{\sqrt{3}}{2}\\ \sin \text{C}=\frac{\text{side opposite to angle C}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\mathrm{k}}{2\mathrm{k}}=\frac{\sqrt{3}}{2}\\ \cos \text{C}=\frac{\text{side adjacent to angle C}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\mathrm{k}}{2\mathrm{k}}=\frac{1}{2}\\ \text{Now,}\\ \text{(i)}\sin \text{A}\cos \text{C}+\cos \text{A}\sin \text{C}=\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=1\\ \text{(ii)}\cos \text{A}\cos \text{C}-\sin \text{A}\sin \text{C}=\frac{\sqrt{3}}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{\sqrt{3}}{2}=0\end{array}$

Q.10

$\begin{array}{l}\text{In }\mathrm{\Delta }\text{\hspace{0.17em}PQR, right-angled at Q, PR}+\text{QR}=\text{25 cm and PQ}=\text{5 cm.}\\ \text{Determine the values of sin\hspace{0.17em}P, cos\hspace{0.17em}P and tan\hspace{0.17em}\hspace{0.17em}P.}\end{array}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{PR}+\text{QR}=\text{25 cm}\\ \text{and}\\ \text{PQ}=\text{5 cm}\\ \text{Let PR be}\mathrm{x}\text{. Then, QR =}25-\mathrm{x}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{\hspace{0.17em}PQR, we get}\\ {\text{PR}}^2={\text{PQ}}^2+{\text{QR}}^2\\ \text{or}{\mathrm{x}}^2={\left(5\right)}^2+{(25-\mathrm{x})}^2\\ \text{or}{\mathrm{x}}^2=25+625+{\mathrm{x}}^2-50\mathrm{x}\\ \text{or \hspace{0.17em}}50\mathrm{x}=650\\ \text{or \hspace{0.17em}}\mathrm{x}=13\\ \therefore \text{PR}=\mathrm{x}=13\text{cm}\\ \text{and}\\ \text{QR}=25-\mathrm{x}=25-13=12\text{cm}\\ \sin \text{\hspace{0.17em}P}=\frac{\text{side opposite to angle P}}{\text{hypotenuse}}=\frac{\text{QR}}{\text{PR}}=\frac{12}{13}\\ \cos \text{\hspace{0.17em}P}=\frac{\text{side adjacent to angle P}}{\text{hypotenuse}}=\frac{\text{PQ}}{\text{PR}}=\frac{5}{13}\\ \tan \text{\hspace{0.17em}P}=\frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}}=\frac{\text{QR}}{\text{PQ}}=\frac{12}{5}\end{array}$

Q.11

$\begin{array}{l}\text{State whether the following are true or false. Justify}\\ \text{your answer.}\\ \text{(i) The value of tan A is always less than 1.}\\ \text{(ii) sec A}=\frac{12}{5}\text{ for some value of angle A.}\\ \text{(iii) cos A is the abbreviation used for the cosecant}\\ \text{of angle A.}\\ \text{(iv) cot A is the product of cot and A.}\\ \text{(v) sin }\mathrm{\theta }=\frac{4}{3}\text{ for some angle }\mathrm{\theta }.\end{array}$

Ans.

(i) Let us consider the following Δ ABC, right-angled at B.

$\begin{array}{l}\\ \tan \text{\hspace{0.17em}A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{12}{5}>1\\ \therefore \tan \text{\hspace{0.17em}A}>1\\ \text{So, tan\hspace{0.17em}A<1 is not always true.}\\ \text{Hence, the given statement is false.}\end{array}$

(ii) Let us consider the following Δ ABC, right-angled at B.

$\begin{array}{l}\text{We consider on the above}\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B.}\\ \\ \sec \text{\hspace{0.17em}A}=\frac{\text{hypotenuse}}{\text{side adjacent to angle A}}=\frac{\text{AC}}{\text{AB}}=\frac{12}{5}\\ \text{Now, by using Pythagoras theorem, we have}\\ {\text{BC}}^2={\text{AC}}^2-{\text{AB}}^2=144-25=119\\ \text{or BC}=\sqrt{119}=10.9\\ \text{Therefore,}\sec \text{\hspace{0.17em}A}=\frac{12}{5}\text{is possible for some values}\\ \text{of angle A.}\\ \text{Hence, the given statement is true.}\\ \text{(iii)}\\ \text{The abbreviation used for the cosecant of angle A is cosec\hspace{0.17em}A.}\\ \text{Therefore, the given statement is false.}\\ \text{(iv)}\\ \text{cot A is not the product of cot and A. Therefore, the given}\\ \text{statement is true.}\\ \text{(v)}\\ \sin \mathrm{\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}\\ \text{In a right-angled triangle hypotenuse is the longest side.}\\ \text{Thus,}\\ \sin \mathrm{\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}<1\\ \text{Therefore,}\sin \mathrm{\theta }=\frac{4}{3}>1\text{is not possible for any value of}\mathrm{\theta }.\\ \text{Hence, the given statement is false.}\end{array}$

Q.12

$\begin{array}{l}\text{Evaluate the following:}\\ \text{(i) sin\hspace{0.17em}60°\hspace{0.17em}cos 30°}+\text{sin 30°\hspace{0.17em}cos 60°}\\ {\text{(ii) 2tan}}^{\text{2}}\text{45°}+{\text{cos}}^{\text{2}}\text{30°}-{\text{sin}}^{\text{2}}\text{60°}\\ \text{(iii)}\frac{\text{cos 45°}}{\text{sec 30°}+\text{cosec 30°}}\\ \text{(iv)}\frac{\text{sin 30°}+\text{tan 45°}-\text{cosec 60°}}{\text{sec 30°}+\text{cos 60°}+\text{cot 45°}}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\frac{{\text{5cos}}^2\text{60°}+{\text{4sec}}^2\text{30°}-{\text{tan}}^2\text{45°}}{{\text{sin}}^2\text{30°}+{\text{cos}}^2\text{30°}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}60°\hspace{0.17em}cos 30°}+\text{sin 30°\hspace{0.17em}cos 60°}=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}=1\\ \text{(ii)}\\ {\text{2tan}}^{\text{2}}\text{45°}+{\text{cos}}^{\text{2}}\text{30°}-{\text{sin}}^{\text{2}}\text{60°}=2\times {\left(1\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2=2\\ \text{(iii)}\\ \frac{\text{cos 45°}}{\text{sec 30°}+\text{cosec 30°}}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3)}}\\ =\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3)}}\times \frac{(1-\sqrt{3)}}{(1-\sqrt{3)}}\\ =\frac{\sqrt{3}-3}{2\sqrt{2}(1-3)}=\frac{3-\sqrt{3}}{4\sqrt{2}}\\ \text{(iv)}\\ \frac{\text{sin 30°}+\text{tan 45°}-\text{cosec 60°}}{\text{sec 30°}+\text{cos 60°}+\text{cot 45°}}=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}\\ =\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\\ =\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \frac{3\sqrt{3}-4}{3\sqrt{3}-4}\\ =\frac{{(3\sqrt{3}-4)}^2}{{\left(3\sqrt{3}\right)}^2-16}=\frac{{\left(3\sqrt{3}\right)}^2-24\sqrt{3}+16}{27-16}\\ =\frac{43-24\sqrt{3}}{11}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\\ \frac{{\text{5cos}}^2\text{60°}+{\text{4sec}}^2\text{30°}-{\text{tan}}^2\text{45°}}{{\text{sin}}^2\text{30°}+{\text{cos}}^2\text{30°}}=\frac{5\times {\left(\frac{1}{2}\right)}^2+4\times {\left(\frac{2}{\sqrt{3}}\right)}^2-1}{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\\ =\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}=\frac{67}{12}\end{array}$

Q.13

$\begin{array}{l}\text{Choose the correct option and justify your choice:}\\ \text{(i)}\frac{\text{2tan30°}}{\text{1}+{\text{tan}}^2\text{30°}}=\\ \text{(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°}\\ \text{(ii)}\frac{\text{1}-{\text{tan}}^2\text{45°}}{\text{1}+{\text{tan}}^2\text{45°}}=\\ \text{(A) tan 90° (B) 1 (C) sin 45° (D) 0}\\ \text{(iii) sin2A}=\text{2sinA is true when A}=\\ \text{(A) 0° (B) 30° (C) 45° (D) 60°}\\ \text{(iv)}\frac{\text{2tan30°}}{1-{\text{tan}}^2\text{30°}}=\\ \text{(A) cos 60° (B)sin 60° (C) tan 60° (D) sin 30}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \frac{\text{2tan30°}}{\text{1}+{\text{tan}}^2\text{30°}}=\frac{2\times \frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}=\frac{2\times 3}{4\sqrt{3}}=\frac{\sqrt{3}}{2}=\sin 60\mathrm{°}\\ \text{Hence, the correct option is (A).}\\ \text{(ii)}\\ \frac{\text{1}-{\text{tan}}^2\text{45°}}{\text{1}+{\text{tan}}^2\text{45°}}=\frac{1-1}{1+1}=0\\ \text{Hence, the correct option is (D).}\\ \text{(iii)}\\ \text{sin2A}=\text{2sinA}\\ \text{On putting A}=0\mathrm{°},\text{we get}\\ \text{sin\hspace{0.17em}}0\mathrm{°}=\text{2sin\hspace{0.17em}}0\mathrm{°}\\ \text{or 0}=2\times 0\\ \text{or 0}=0\\ \text{Therefore, the correct option is (A).}\\ \text{Other given options does not satisfy the given condition.}\\ \text{(iv)}\\ \frac{\text{2tan30°}}{1-{\text{tan}}^2\text{30°}}=\frac{\text{2}\times \frac{1}{\sqrt{3}}}{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3}=\tan 60\mathrm{°}\\ \text{Therefore, the correct option is (C).}\end{array}$

Q.14

$\text{If tan}(\mathrm{A}+\mathrm{B})=\sqrt{\text{3}}\text{ and tan}(\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{\text{3}}}\text{; 0° < A + B}\le \text{90°; A > B, find A and B.}$

Ans.

$\begin{array}{l}\text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan}(\mathrm{A}+\mathrm{B})=\sqrt{\text{3}}\\ \text{or tan (A}+\text{B})=\text{tan\hspace{0.17em}60°}\\ \text{or A}+\text{B}=60\mathrm{°}...\text{(1)}\\ \text{Again, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan}(\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{\text{3}}}\\ \text{or tan (A}-\text{B})=\text{tan\hspace{0.17em}}3\text{0°}\\ \text{or A}-\text{B}=30\mathrm{°}...\text{(2)}\\ \text{On adding equations (1) and (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 2A}=90\mathrm{°}\\ \text{or A}=45\mathrm{°}\\ \text{On putting this value of A in equation (1), we get}\\ \text{B}=15\mathrm{°}\end{array}$

Q.15

$\begin{array}{l}\text{State whether the following are true or false.}\\ \text{Justify your answer.}\\ \text{(i) sin}(\mathrm{A}+\mathrm{B})=\text{sin A}+\text{sin B.}\\ \text{(ii) The value of sin}\mathrm{\theta }\text{ increases as }\mathrm{\theta }\text{ increases.}\\ \text{(iii) The value of cos}\mathrm{\theta }\text{ increases as }\mathrm{\theta }\text{ increases.}\\ \text{(iv) sin}\mathrm{\theta }=\text{cos}\mathrm{\theta }\text{ for all values of }\mathrm{\theta }.\\ \text{(v) cot A is not defined for A}=0.\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Let A}=30\mathrm{°}\text{and B}=60\mathrm{°}.\\ \text{Now,}\\ \sin (\text{A}+\text{B)}=\sin (30\mathrm{°}+60\mathrm{°}\text{)}=\sin 90\mathrm{°}=1\\ \text{and}\\ \text{sin A}+\text{sin B}=\text{sin}30\mathrm{°}+\text{sin\hspace{0.17em}\hspace{0.17em}}60\mathrm{°}=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\\ \therefore \sin (\text{A}+\text{B)}\ne \text{sin A}+\text{sin B}\\ \text{Hence, the given statement is false.}\\ \text{(ii)}\\ \text{We know that}\\ \sin 0\mathrm{°}=0,\\ \sin 30\mathrm{°}=\frac{1}{2}=0.5,\\ \sin 45\mathrm{°}=\frac{1}{\sqrt{2}}=0.7,\\ \sin 60\mathrm{°}=\frac{\sqrt{3}}{2}=0.87,\\ \sin 90\mathrm{°}=1\\ \text{We see that the value of sin}\mathrm{\theta }\text{increases as}\mathrm{\theta }\text{increases}\\ \text{in the interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{Hence, the given statement is true if}\mathrm{\theta }\text{lies in the}\\ \text{interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{(iii)}\\ \text{We know that}\\ \cos 0\mathrm{°}=1,\\ \cos 30\mathrm{°}=\frac{\sqrt{3}}{2}=0.87,\\ \cos 45\mathrm{°}=\frac{1}{\sqrt{2}}=0.7,\\ \cos 60\mathrm{°}=\frac{1}{2}=0.5,\\ \cos 90\mathrm{°}=0\\ \text{We see that the value of}\mathrm{cos\theta }\text{decreases as}\mathrm{\theta }\text{increases}\\ \text{in the interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{Hence, the given statement is false if}\mathrm{\theta }\text{lies in the}\\ \text{interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{(iv)}\\ \text{We know that}\\ \sin 0\mathrm{°}=0\text{and}\cos 0\mathrm{°}=1,\\ \sin 30\mathrm{°}=\frac{1}{2}=0.5\text{and}\cos 30\mathrm{°}=\frac{\sqrt{3}}{2}=0.87.\\ \text{Therefore, sin}\mathrm{\theta }\ne \text{cos}\mathrm{\theta }\text{for all values of}\mathrm{\theta }.\\ \text{Hence, the given statement is false}.\\ \text{(v)}\\ \text{We know that cot A}=\frac{\cos \text{A}}{\sin \text{A}}.\\ \text{So, for A}=0,\text{we have}\\ \text{cot 0}=\frac{\cos \text{0}}{\sin \text{0}}=\frac{1}{0}\\ \text{We know that division by 0 is not defined. Therefore, cot 0}\\ \text{is not defined.}\\ \text{Hence, the given statement is true}.\end{array}$

Q.16

$\begin{array}{l}\text{Evaluate\hspace{0.17em}:}\\ \text{(i)}\frac{\text{sin 18°}}{\text{cos 72}°}\\ \text{(ii)}\frac{\text{tan 26°}}{\text{cot 64°}}\\ \text{(iii) cos 48°}-\text{sin 42°}\\ \text{(iv) cosec 31°}-\text{sec 59°}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \frac{\text{sin 18°}}{\text{cos 72}\mathrm{°}}=\frac{\text{cos(90°}-18\text{°})}{\text{cos 72}\mathrm{°}}=\frac{\text{cos 72}\mathrm{°}}{\text{cos 72}\mathrm{°}}=1\\ \\ \text{(ii)}\\ \frac{\text{tan 26°}}{\text{cot 64°}}=\frac{\text{cot (90°}-\text{26°)}}{\text{cot 64°}}=\frac{\text{cot 64°}}{\text{cot 64°}}=1\\ \\ \text{(iii)}\\ \text{cos 48°}-\text{sin 42°}=\text{sin (90°}-\text{48°)}-\text{sin 42°}\\ =\text{sin 42°}-\text{sin 42°}\\ =0\\ \text{(iv)}\\ \text{cosec 31°}-\text{sec 59°}=\text{sec (90°}-\text{31°)}-\text{sec 59°}\\ =\text{sec 59°}-\text{sec 59°}\\ =0\end{array}$

Q.17

$\begin{array}{l}\text{Show that\hspace{0.17em}:}\\ \text{(i) tan 48° tan 23° tan 42° tan 67°}=1\\ \text{(ii) cos 38° cos 52°}-\text{sin 38° sin 52°}=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{tan 48° tan 23° tan 42° tan 67°}\\ =\text{cot (90°}-\text{48°) cot (90°}-23\text{°) tan 42° tan 67°}\\ =\text{cot}42°\text{\hspace{0.17em} cot 67°\hspace{0.17em}\hspace{0.17em}tan 42°\hspace{0.17em}\hspace{0.17em}tan 67°}\\ =\text{cot}42°\text{\hspace{0.17em}tan 42°\hspace{0.17em}\hspace{0.17em}tan 67° cot 67°}\\ =1\times 1=1\\ \text{(ii)}\\ \text{cos 38° cos 52°}-\text{sin 38° sin 52°}\\ =\sin \text{(90°}-3\text{8°)\hspace{0.17em}cos (90°}-3\text{8°)}-\text{sin 38° sin 52°}\\ =\sin \text{\hspace{0.17em}\hspace{0.17em}52°\hspace{0.17em}}\sin 38\text{°}-\text{sin 38° sin 52°}\\ =0\end{array}$

Q.18

$\text{If tan 2A}=\text{cot}(\text{A}-18\mathrm{°}),\text{where 2A is an acute angle},\text{find the value of A}.$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{tan 2A}=\text{cot}(\text{A}-18\mathrm{°})\\ \text{or cot (90°}-2\text{A})=\text{cot}(\text{A}-18\mathrm{°})\\ \text{or 90°}-2\text{A}=\text{A}-18\mathrm{°}\\ \text{or}3\text{A}=108\mathrm{°}\\ \text{or A}=\frac{108\mathrm{°}}{3}=36\mathrm{°}\end{array}$

Q.19

$\text{If tan A}=\text{cot B, prove that A}+\text{B}=\text{90°.}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan A}=\text{cot B}\\ \text{or cot (90°}-\text{A)}=\cot \text{\hspace{0.17em}\hspace{0.17em}B}\\ \text{or 90°}-\text{A}=\text{B}\\ \text{or A}+\text{B}=\text{90°}\end{array}$

Q.20

$\text{If sec 4A}=\text{cosec (A}-\text{20°), where 4A is an acute angle, find the value of A.}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{sec 4A}=\text{cosec (A}-\text{20°)}\\ \text{or cosec (90°}-\text{4A)}=\text{cosec (A}-\text{20°)}\\ \text{or 90°}-\text{4A}=\text{A}-\text{20°}\\ \text{or 5A}=110\text{°}\\ \text{or A}=\frac{110\text{°}}{5}=22\text{°}\end{array}$

Q.21

$\begin{array}{l}\text{If A, B and C are interior angles of a triangle ABC,}\\ \text{then show that}\\ \text{sin}\left(\frac{\text{B+C}}{2}\right)=\cos \frac{\text{A}}{2}.\end{array}$

Ans.

$\begin{array}{l}\text{Given that A, B and C are interior angles of a triangle ABC.}\\ \therefore \text{A}+\text{B}+\text{C}=180\mathrm{°}\\ \text{or A}=180\mathrm{°}-\text{B}-\text{C}\\ \text{Now,}\\ \text{sin}\left(\frac{\text{B}+\text{C}}{2}\right)=\cos (90\mathrm{°}-\frac{\text{B}+\text{C}}{2})\\ =\cos \left(\frac{180\mathrm{°}-\text{B}-\text{C}}{2}\right)\\ =\cos \left(\frac{\text{A}}{2}\right)\end{array}$

Q.22

$\text{Express sin 67°}+\text{cos 75° in terms of trigonometric ratios of angles between 0° and 45°.}$

Ans.

$\begin{array}{l}\text{sin 67°}+\text{cos 75°}=\cos \text{(90°}-\text{67°)}+\text{sin (90°}-\text{75°)}\\ =\cos 23\text{°}+\text{sin 15°}\end{array}$

Q.23

$\text{Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.}$

Ans.

$\begin{array}{l}\text{We know that}\\ {\text{cosec}}^{\text{2}}\text{\hspace{0.17em}A}=1+{\text{cot}}^2\text{\hspace{0.17em}A}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\sin }^2\text{A}}=1+{\text{cot}}^2\text{\hspace{0.17em}A}\\ \text{or \hspace{0.17em}\hspace{0.17em}}{\sin }^2\text{A}=\frac{1}{1+{\text{cot}}^2\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}}\sin \text{A}=\frac{1}{\sqrt{1+{\text{cot}}^2\text{\hspace{0.17em}A}}}\\ \text{Also, we know that}\\ {\text{sec}}^{\text{2}}\text{\hspace{0.17em}A}=1+{\text{tan}}^2\text{\hspace{0.17em}A}\\ {\text{or sec}}^{\text{2}}\text{\hspace{0.17em}A}=1+\frac{1}{{\text{cot}}^2\text{\hspace{0.17em}A}}\\ \text{or}\sec \text{\hspace{0.17em}A}=\sqrt{\frac{1+{\cot }^2\text{A}}{{\cot }^2\text{A}}}=\frac{\sqrt{1+{\cot }^2\text{A}}}{\cot \text{A}}\\ \text{Also, we know that}\\ \text{tan\hspace{0.17em}A}=\frac{\sin \text{A}}{\cos \text{A}}=\frac{1}{\frac{\cos \text{A}}{\sin \text{A}}}=\frac{1}{\cot \text{A}}\end{array}$

Q.24

$\text{Write all the other trigonometric ratios of}\angle \text{A in terms of sec A.}$

Ans.

$\begin{array}{l}\text{We know that}\\ {\text{sin}}^{\text{2}}\text{\hspace{0.17em}A}=1-{\text{cos}}^2\text{\hspace{0.17em}A}\\ {\text{or \hspace{0.17em}\hspace{0.17em}sin}}^{\text{2}}\text{\hspace{0.17em}A}=1-\frac{1}{{\text{sec}}^2\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}\hspace{0.17em}}{\sin }^2\text{A}=\frac{{\text{sec}}^2\text{\hspace{0.17em}A}-1}{{\text{sec}}^2\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}}\sin \text{A}=\frac{\sqrt{{\text{sec}}^2\text{\hspace{0.17em}A}-1}}{\text{sec\hspace{0.17em}A}}\\ \mathrm{or}\frac{1}{\mathrm{cosec}\mathrm{A}}=\frac{\sqrt{{\text{sec}}^2\text{\hspace{0.17em}A}-1}}{\text{sec\hspace{0.17em}A}}\\ \mathrm{or}\mathrm{cosec}\mathrm{A}=\frac{\text{sec\hspace{0.17em}A}}{\sqrt{{\text{sec}}^2\text{\hspace{0.17em}A}-1}}\\ \text{Also, we know that}\\ \text{sec\hspace{0.17em}A cos\hspace{0.17em}A}=1\\ \text{or cos\hspace{0.17em}A}=\frac{1}{\text{sec\hspace{0.17em}A}}\\ \\ \text{Also, we know that}\\ {\text{sec}}^2\text{A}-{\text{tan}}^2\text{\hspace{0.17em}A}=1\\ {\text{or tan}}^2\text{\hspace{0.17em}A}={\text{sec}}^2\text{A}-1\\ \mathrm{or}\text{tan\hspace{0.17em}A}=\sqrt{{\text{sec}}^2\text{A}-1}\\ \text{Also, we know that}\\ \text{tan\hspace{0.17em}A cot\hspace{0.17em}A}=1\\ \mathrm{or}\text{cot\hspace{0.17em}A}=\frac{1}{\text{tan\hspace{0.17em}A}}=\frac{1}{\sqrt{{\text{sec}}^2\text{A}-1}}\end{array}$

Q.25

$\begin{array}{l}\text{Evaluate:}\\ \text{(i)}\frac{{\text{sin}}^2\text{63°}+{\text{sin}}^2\text{27°}}{{\text{cos}}^2\text{17°}+{\text{cos}}^2\text{73°}}\\ \text{(ii) sin 25° cos 65°}+\text{cos 25° sin 65°}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \frac{{\text{sin}}^2\text{63°}+{\text{sin}}^2\text{27°}}{{\text{cos}}^2\text{17°}+{\text{cos}}^2\text{73°}}=\frac{{\text{sin}}^2\text{\hspace{0.17em}63°}+\mathrm{co}{\text{s}}^2\text{(90°}-\text{27°)}}{{\text{sin}}^2\text{(90°}-\text{17°)}+{\text{cos}}^2\text{\hspace{0.17em}73°}}\\ =\frac{{\text{sin}}^2\text{\hspace{0.17em}63°}+\mathrm{co}{\text{s}}^2\text{\hspace{0.17em}63°}}{{\text{sin}}^2\text{\hspace{0.17em}73°}+{\text{cos}}^2\text{\hspace{0.17em}73°}}\\ =\frac{1}{1}=1\\ \text{(ii)}\\ \text{sin\hspace{0.17em}25° cos 65°}+\text{cos\hspace{0.17em}25° sin\hspace{0.17em}65°}\\ =\text{sin\hspace{0.17em}25°\hspace{0.17em}sin(90°}-\text{65°})+\text{cos 25° cos(90°}-\text{\hspace{0.17em}65°)}\\ =\text{sin\hspace{0.17em}25° sin\hspace{0.17em}25°}+\text{cos 25°\hspace{0.17em}cos 25°}\\ ={\text{sin}}^2\text{\hspace{0.17em}25°}+{\text{cos}}^2\text{25°}\\ =1\end{array}$

Q.26

$\begin{array}{l}\text{Choose the correct option. Justify your choice.}\\ {\text{(i) 9 sec}}^2\text{A}-{\text{9 tan}}^2\text{A}=\\ \text{(A) 1 (B) 9 (C) 8 (D) 0}\\ \text{(ii) (1 + tan}\mathrm{\theta }+\text{sec}\mathrm{\theta }\text{) (1 + cot}\mathrm{\theta }-\text{cosec}\mathrm{\theta }\text{)}=\\ \text{(A) 0 (B) 1 (C) 2 (D)}-\text{1}\\ \text{(iii) (sec A + tan A) (1}-\text{sin A)}=\\ \text{(A) sec A (B) sin A (C) cosec A (D) cos A}\\ \text{(iv)}\frac{1+{\tan }^2\mathrm{A}}{1+{\cot }^2\mathrm{A}}=\\ {\text{(A) sec}}^2\mathrm{A}\text{(B)}-{\text{1 (C) cot}}^2{\text{A (D) tan}}^2\mathrm{A}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ {\text{9 sec}}^2\text{A}-{\text{9 tan}}^2\text{A}=9({\text{sec}}^2\text{A}-{\text{tan}}^2\text{A)}=9\times 1=9\\ \text{Therefore, the correct option is (B).}\\ \text{(ii)}\\ \text{(1}+\text{tan}\mathrm{\theta }+\text{sec}\mathrm{\theta }\text{) (}1+\mathrm{cot\theta }-\mathrm{cosec}\mathrm{\theta }\text{)}\\ =(1+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{1}{\mathrm{cos\theta }})(1+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}-\frac{1}{\mathrm{sin\theta }})\\ =\left(\frac{\mathrm{sin\theta }+\mathrm{cos\theta }+1}{\mathrm{cos\theta }}\right)\left(\frac{\mathrm{sin\theta }+\mathrm{cos\theta }-1}{\mathrm{sin\theta }}\right)\\ =\frac{{(\mathrm{sin\theta }+\mathrm{cos\theta })}^2-1}{\mathrm{sin\theta cos\theta }}=\frac{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }+2\mathrm{sin\theta cos\theta }-1}{\mathrm{sin\theta cos\theta }}\\ =\frac{1+2\mathrm{sin\theta cos\theta }-1}{\mathrm{sin\theta cos\theta }}=2\\ \text{Therefore, the correct option is (C).}\\ \text{(iii)}\\ \text{(sec A + tan A) (1}-\text{sin A)}\\ =(\frac{1}{\text{cos A}}+\frac{\sin \text{A}}{\text{cos A}})(\text{1}-\text{sin A})\\ =\left(\frac{1+\sin \text{A}}{\text{cos A}}\right)(\text{1}-\text{sin A})\\ =\frac{1-{\sin }^2\text{A}}{\text{cos A}}=\frac{{\cos }^2\text{A}}{\text{cos A}}=\cos \text{A}\\ \text{Therefore, the correct option is (D).}\\ \text{(iv)}\\ \frac{1+{\tan }^2\text{A}}{1+{\cot }^2\text{A}}=\frac{{\sec }^2\text{A}}{{\mathrm{cosec}}^2\text{A}}=\frac{{\sin }^2\text{A}}{{\cos }^2\text{A}}={\tan }^2\text{A}\\ \text{Therefore, the correct option is (D).}\end{array}$

Q.27

$\begin{array}{l}\text{Prove the following identities, where the angles involved}\\ \text{are acute angles for which the expressions are defined.}\\ \text{(i) (cosec}\mathrm{\theta }-\text{cot}\mathrm{\theta }{)}^2=\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}\\ \text{(ii)}\frac{\cos \text{A}}{1+\sin \text{A}}+\frac{1+\sin \text{A}}{\cos \text{A}}=2\sec \text{A}\\ \text{(iii)}\frac{\mathrm{tan\theta }}{1-\mathrm{cot\theta }}+\frac{\mathrm{cot\theta }}{1-\mathrm{tan\theta }}=1+\mathrm{sec\theta cosec\theta }\\ \text{[Hint :Write the expression in terms of sin}\mathrm{\theta }\text{and cos}\mathrm{\theta }.]\\ \text{(iv)}\frac{1+\sec \text{A}}{\sec \text{A}}=\frac{{\sin }^2\text{A}}{1-\cos \text{A}}\\ \text{[Hint: Simplify LHS and RHS separately]}\\ \text{(v)}\frac{\cos \text{A}-\sin \text{A}+1}{\cos \text{A}+\sin \text{A}-1}=\text{cosec A}+\text{cot A, using the identity}\\ {\text{cosec}}^{\text{2}}\text{A}=\text{1}+{\text{cot}}^2\text{\hspace{0.17em}A.}\\ \text{(vi)}\sqrt{\frac{1+\sin \text{A}}{1-\sin \text{A}}}=\sec \text{A}+\tan \text{A}\\ \text{(vii)}\frac{\text{sin}\mathrm{\theta }-{\text{2sin}}^3\mathrm{\theta }}{{\text{2cos}}^3\mathrm{\theta }-\text{cos}\mathrm{\theta }}=\text{tan}\mathrm{\theta }\\ {\text{(viii) (sin A + cosec A)}}^2+{(\text{cos A}+\text{sec A})}^{\text{2}}=7+{\tan }^2\text{A}+{\cot }^2\text{A}\\ \text{(ix) (cosec A}-\text{sin A) (sec A}-\text{cos A)}=\frac{1}{\tan \text{A}+\cot \text{A}}\\ \text{[Hint: Simplify LHS and RHS separately]}\\ \text{(x)}\frac{1+{\tan }^2\text{A}}{1+{\cot }^2\text{A}}={\left(\frac{\text{1}-\text{tanA}}{\text{1}-\text{cot A}}\right)}^2={\tan }^2\text{A}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{LHS}={(\text{cosec}\mathrm{\theta }-\text{cot}\mathrm{\theta })}^2\\ ={\mathrm{cosec}}^2\mathrm{\theta }+{\cot }^2\mathrm{\theta }-2\text{cosec}\mathrm{\theta }\text{\hspace{0.17em}cot}\mathrm{\theta }\\ =\frac{1}{{\sin }^2\mathrm{\theta }}+\frac{{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}-\frac{2\mathrm{cos\theta }}{{\sin }^2\mathrm{\theta }}\\ =\frac{1+{\cos }^2\mathrm{\theta }-2\mathrm{cos\theta }}{{\sin }^2\mathrm{\theta }}\\ =\frac{(1-\mathrm{cos\theta })(1-\mathrm{cos\theta })}{1-{\cos }^2\mathrm{\theta }}=\frac{(1-\mathrm{cos\theta })(1-\mathrm{cos\theta })}{(1+\mathrm{cos\theta })(1-\mathrm{cos\theta })}\\ =\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}=\text{RHS}\\ \\ \text{(ii)}\\ \text{LHS}=\frac{\cos \text{A}}{1+\sin \text{A}}+\frac{1+\sin \text{A}}{\cos \text{A}}\\ =\frac{{\cos }^2\text{A}+{(1+\sin \text{A)}}^2}{(1+\sin \text{A)}\cos \text{A}}\\ =\frac{{\cos }^2\text{A}+{\sin }^2\text{A}+2\sin \text{A}+1}{(1+\sin \text{A)}\cos \text{A}}\\ =\frac{1+1+2\sin \text{A}}{(1+\sin \text{A)}\cos \text{A}}\\ =\frac{2(1+\sin \text{A)}}{(1+\sin \text{A)}\cos \text{A}}=2\sec \text{A}=\text{RHS}\\ \\ \text{(iii)}\\ \text{LHS}=\frac{\mathrm{tan\theta }}{1-\mathrm{cot\theta }}+\frac{\mathrm{cot\theta }}{1-\mathrm{tan\theta }}\\ \\ =\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{1-\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}+\frac{\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}{1-\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}\\ =\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{\frac{\mathrm{sin\theta }-\mathrm{cos\theta }}{\mathrm{sin\theta }}}+\frac{\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}{\frac{\mathrm{cos\theta }-\mathrm{sin\theta }}{\mathrm{cos\theta }}}\\ =\frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }(\mathrm{sin\theta }-\mathrm{cos\theta })}+\frac{{\cos }^2\mathrm{\theta }}{\mathrm{sin\theta }(\mathrm{cos\theta }-\mathrm{sin\theta })}\\ =\frac{{\sin }^3\mathrm{\theta }-{\cos }^3\mathrm{\theta }}{\mathrm{sin\theta cos\theta }(\mathrm{sin\theta }-\mathrm{cos\theta })}\\ =\frac{({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }+\mathrm{sin\theta cos\theta })(\mathrm{sin\theta }-\mathrm{cos\theta })}{\mathrm{sin\theta cos\theta }(\mathrm{sin\theta }-\mathrm{cos\theta })}\\ =\frac{1+\mathrm{sin\theta cos\theta }}{\mathrm{sin\theta cos\theta }}=1+\mathrm{sec\theta cosec}\mathrm{\theta }=\text{RHS}\\ \text{(iv)}\\ \text{LHS}=\frac{1+\sec \text{A}}{\sec \text{A}}=\frac{1+\frac{1}{\cos \text{A}}}{\frac{1}{\cos \text{A}}}=\frac{1+\cos \text{A}}{1}\times \frac{1-\cos \text{A}}{1-\cos \text{A}}\\ =\frac{1-{\cos }^2\text{A}}{1-\cos \text{A}}=\frac{{\sin }^2\text{A}}{1-\cos \text{A}}=\text{RHS}\\ \text{(v)}\\ \text{LHS}=\frac{\cos \text{A}-\sin \text{A}+1}{\cos \text{A}+\sin \text{A}-1}\\ =\frac{\frac{\cos \text{A}}{\sin \text{A}}-\frac{\sin \text{A}}{\sin \text{A}}+\frac{1}{\sin \text{A}}}{\frac{\cos \text{A}}{\sin \text{A}}+\frac{\sin \text{A}}{\sin \text{A}}-\frac{1}{\sin \text{A}}}\\ =\frac{\cot \text{A}-1+\mathrm{cosec}\text{A}}{\cot \text{A}+1-\mathrm{cosec}\text{\hspace{0.17em}A}}\\ =\frac{\cot \text{A}-(1-\mathrm{cosec}\text{A)}}{\cot \text{A}+(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}\times \frac{\cot \text{A}-(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}{\cot \text{A}-(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}\\ =\frac{{\cot }^2\text{A}+{(1-\mathrm{cosec}\text{A)}}^2-2\mathrm{cotA}(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}{{\cot }^2\text{A}-{(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}^2}\\ =\frac{{\cot }^2\text{A}+1+{\mathrm{cosec}}^2\text{A}-\text{2}\mathrm{cosec}\text{A}-2\mathrm{cotA}+2\mathrm{cotAcosecA}}{{\cot }^2\text{A}-1-{\mathrm{cosec}}^2\text{\hspace{0.17em}A}+2\mathrm{cosec}\mathrm{A}}\\ =\frac{2{\mathrm{cosec}}^2\text{A}-\text{2}\mathrm{cosec}\text{A}-2\mathrm{cotA}+2\mathrm{cotAcosecA}}{-1-1+2\mathrm{cosec}\mathrm{A}}\\ =\frac{\mathrm{cosec}\text{A(}\mathrm{cosec}\text{A}-\text{1})+\mathrm{cotA}(\mathrm{cosec}\mathrm{A}-1)}{\mathrm{cosec}\text{\hspace{0.17em}A}-1}\\ =\mathrm{cosec}\text{A}+\text{cot\hspace{0.17em}A}\\ \text{(vi)}\\ \text{LHS}=\sqrt{\frac{1+\sin \text{A}}{1-\sin \text{A}}}=\sqrt{\frac{1+\sin \text{A}}{1-\sin \text{A}}\times \frac{1+\sin \text{A}}{1+\sin \text{A}}}\\ =(1+\sin \text{A})\sqrt{\frac{1}{1-{\sin }^2\text{A}}}=\frac{1+\sin \text{A}}{\cos \text{A}}=\mathrm{tanA}+\mathrm{secA}=\text{RHS}\\ \text{(vii)}\\ \text{LHS}=\frac{\text{sin}\mathrm{\theta }-{\text{2sin}}^3\mathrm{\theta }}{{\text{2cos}}^3\mathrm{\theta }-\text{cos}\mathrm{\theta }}\\ =\frac{\text{sin}\mathrm{\theta }(1-{\text{2sin}}^2\mathrm{\theta })}{\text{cos}\mathrm{\theta }({\text{2cos}}^2\mathrm{\theta }-1)}\\ =\frac{\text{sin}\mathrm{\theta }(1-{\text{2sin}}^2\mathrm{\theta })}{\text{cos}\mathrm{\theta }\left(\text{2(1}-{\sin }^2\mathrm{\theta })-1\right)}\\ =\frac{\text{sin}\mathrm{\theta }(1-{\text{2sin}}^2\mathrm{\theta })}{\text{cos}\mathrm{\theta }(\text{2}-2{\sin }^2\mathrm{\theta }-1)}\\ =\frac{\text{sin}\mathrm{\theta }(1-{\text{2sin}}^2\mathrm{\theta })}{\text{cos}\mathrm{\theta }(1-2{\sin }^2\mathrm{\theta })}\\ =\mathrm{tan\theta }=\mathrm{RHS}\\ \text{(viii)}\\ \text{LHS}={(\text{sin A}+\text{cosec A})}^2+{(\text{cos A}+\text{sec A})}^{\text{2}}\\ ={\sin }^2\mathrm{A}+{\mathrm{cosec}}^2\mathrm{A}+2+{\cos }^2\mathrm{A}+{\sec }^2\mathrm{A}+2\\ ={\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}+4+{\sec }^2\mathrm{A}+{\mathrm{cosec}}^2\mathrm{A}\\ =1+4+1+{\tan }^2\mathrm{A}+1+{\cot }^2\mathrm{A}\\ =7+{\tan }^2\mathrm{A}+{\cot }^2\mathrm{A}=\text{RHS}\\ \text{(ix)}\\ \text{LHS}=\text{(cosec A}-\text{sin A) (sec A}-\text{cos A)}\\ =(\frac{1}{\mathrm{sinA}}-\mathrm{sinA})(\frac{1}{\mathrm{cosA}}-\mathrm{cosA})\\ =\left(\frac{1-{\sin }^2\mathrm{A}}{\mathrm{sinA}}\right)\left(\frac{1-{\cos }^2\mathrm{A}}{\mathrm{cosA}}\right)\\ =\frac{{\cos }^2\mathrm{A}}{\mathrm{sinA}}\times \frac{{\sin }^2\mathrm{A}}{\mathrm{cosA}}=\frac{\mathrm{sinAcosA}}{1}=\frac{\mathrm{sinAcosA}}{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}}\\ =\frac{1}{\frac{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}}{\mathrm{sinAcosA}}}=\frac{1}{\frac{\mathrm{sinA}}{\mathrm{cosA}}+\frac{\mathrm{cosA}}{\mathrm{sinA}}}\\ =\frac{1}{\mathrm{tanA}+\mathrm{cotA}}\\ =\mathrm{RHS}\\ \text{(x)}\\ \text{LHS}=\frac{1+{\tan }^2\text{A}}{1+{\cot }^2\text{A}}=\frac{{\sec }^2\mathrm{A}}{{\mathrm{cosec}}^2\mathrm{A}}=\frac{\frac{1}{{\cos }^2\mathrm{A}}}{\frac{1}{{\sin }^2\mathrm{A}}}=\frac{{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}}={\tan }^2\mathrm{A}=\mathrm{RHS}\\ \text{LHS}={\left(\frac{\text{1}-\text{tanA}}{\text{1}-\text{cot A}}\right)}^2={\left(\frac{1-\frac{\mathrm{sinA}}{\mathrm{cosA}}}{1-\frac{\mathrm{cosA}}{\mathrm{sinA}}}\right)}^2={\left(\frac{\frac{\mathrm{cosA}-\mathrm{sinA}}{\mathrm{cosA}}}{\frac{\mathrm{sinA}-\mathrm{cosA}}{\mathrm{sinA}}}\right)}^2\\ =\frac{{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}}={\tan }^2\mathrm{A}=\mathrm{RHS}\end{array}$