CBSE Class 12 Maths Revision Notes Chapter 2

Mathematics is a subject that demands lots of practice. Students need to have an in-depth knowledge of all basic concepts to understand more complex topics of the subject. The Class 12 Mathematics Chapter 2 notes introduce various concepts of the inverse of trigonometric functions. This chapter is very important and has high weightage in the board examinations. These concepts are also to be used to solve MCQs asked in JEE Mains and other national-level competitive examinations.

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Key Topics Covered In Class 12 Mathematics Chapter 2 Notes

The key topics covered in class 12 Mathematics chapter 2 notes include the following.

Inverse function:

Let x and y be two functions such that y = f(x) and x= g(y). Composition of function fog(y) = f(g(y)) and gof(y) = g(f(y))=x then f and y are invertible or inverse of each other, i.e., g = f-1. We can say that, if y=f(x) then x = f-1(y).

Inverse trigonometric function:

If y = f(x) = sin x then its inverse x = sin-1(y), where y ∈ [–2 , 2 ] and x ∈ [-1,1]. This is applicable for all trigonometric functions.

Domain, Range and Graphs of all Inverse trigonometric functions:

Functions	Domain	Range
x= sin (y) y = sin-1(x)	[-1, 1]	[–2 , 2 ]
x= cos (y) y = cos-1(x)	[-1, 1]	[0, ]
x= tan (y) y = tan-1(x)	[∞, -∞] or R	[–2 , 2 ]
x= cot (y) y = cot-1(x)	[∞, -∞] or R	[0, ]
x= sec (y) y = sec-1(x)	R – [-1, 1]	[0, ] – {2}
x= cosec (y) y = cosec-1(x)	R – [-1, 1]	[–2 , 2 ] – {0}

Graphs of Inverse Trigonometric functions:

1. y= sin-1(x)

2. y= cos-1(x)

y = tan-1(x)

y= cot-1(x)

y = cosec-1(x)

y= sec-1(x)

NOTE:

sin-1(x) and tan-1(x) are increasing functions, whereas cos-1(x) and cot-1(x) are decreasing functions over their domain.
sin-1(x) and (sin (x))-1 are different and should not be confused.

Properties:

Property 1: Relation between two trigonometric functions.

sin -1(1x)= cosec-1(x) values of x ∈(−∞,1] ∪ [1,∞)
cos -1(1x)= sec-1(x) values of x ∈ (−∞,1] ∪ [1,∞)
tan -1(1x)= cot-1(x) values of x > 0

= – + cot-1(x) values of x < 0

Property 2: Negative angle Inverse Trigonometric identities

sin -1(x)= -sin-1(x) values of x ∈ [-1, 1]
tan -1(x)= -tan-1(x) values of x ∈ R
cosec -1(x)= -cosec-1(x) values of x ∈ (−∞,−1]∪[1,∞)
cos -1(x)= -cos-1(x) values of x ∈ [-1, 1]
sec -1(x)= -sec-1(x) values of x ∈ (−∞,−1]∪[1,∞)
cot -1(x)= -cot-1(x) values of x ∈ R

Property 3:

sin -1(x)+ cos -1(x)= 2 values of x ∈ [-1, 1]
tan -1(x)+ cot -1(x)= 2 values of x ∈ R
sec -1(x)+ cosec -1(x)= 2 values of x ∈ (−∞,−1]∪[1,∞)

Property 4:

tan -1(x) + tan -1(y)= tan -1x + y1-xy values of xy < 1
tan -1(x) – tan -1(y)= tan -1x – y1+xy values of xy > -1
tan -1(x) – tan -1(y)= + tan -1(x + y1-xy) values of xy > 1; x,y= 0

Property 5:

2tan -1(x)= sin -1(2x1+x2), x 1
2tan -1(x)= cos -1(1-x21+x2), x 0
2tan -1(x)= tan -1(2x1-x2), -1x 1

Property 6:

sin(sin -1(x)) = x, –2 x2

= -x, 2 x32

cos(cos -1(x)) = x, 0 x

= 2-x, x 2

tan(tan -1(x)) = –-x, x [–32 ,- 2 ]

= x, x [–2 , 2 ]

= x – , x [2 ,32 ]

= x – 2, x [32 ,52 ]

Additional Formulas:

sin -1(x) + sin -1(y) = sin -1(x1- y2 +y1- x2 )
sin -1(x) – sin -1(y) = sin -1(x1- y2 -y1- x2 )
cos -1(x) + cos -1(y) = cos -1(xy –1- y2 1- x2 )
cos -1(x) – cos -1(y) = cos -1(xy +1- y2 1- x2 )
tan -1(x) + tan -1(y) + tan -1(z)= tan -1x + y + z – xyz1 – xy – yz – xz , if x, y, z>0 & xy+ yz+ zx<1
If tan -1(x) + tan -1(y) + tan -1(z)= , then x+ y+ z= xyz
If tan -1(x) + tan -1(y) + tan -1(z)= 2, then xy+ yz+ z
sin -1(x) + sin -1(y)+ sin -1(z)= 32, x= y= z= 1
cos -1(x) + cos -1(y)+ cos -1(z)= 3, x= y= z= -1

Remember:

When approaching the equations, square them so that it becomes easier to solve. Sometimes the answers or roots of the equation will not satisfy the original equation.
Do not cancel common factors involving unknown angles on the left-hand side (LHS) and righthand side (RHS) of an equation because it may be the solution of the given equation.
The solution of any equation, including sec θ or tan θ, can never be in the form (2n + 1) π / 2.
The solution of any equation, including cosec θ or cot θ, can never be in the form θ = nπ.

Class 12 Mathematics Chapter 2 Notes Exercises & Answer Solutions.

Chapter 2 of mathematics class 12 helps students to learn the inverse trigonometric functions, their range domain and graphs. It is an important chapter and holds relevance in the 12th board as well as in various entrance exams like JEE, State Engineering Entrance Exams, etc. Extramarks has prepared class 12 mathematics chapter 2 notes which include all Exercise and Answer Solutions. Students can refer to the solutions provided for exercise questions and seek help if they face any difficulties or doubts. The CBSE revision notes include detailed information about theorems, formulas, and derivations in a very descriptive way.

The class 12 mathematics chapter 2 notes are prepared by experts with the objective to provide in-depth knowledge of all the concepts. Practising questions with the help of these solutions will also help to develop analytical and problem-solving skills.

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NCERT Exemplar Class 12 Mathematics

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Key Features of Class 12 Mathematics Chapter 2 Notes

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Q.1

$If \sin^{- 1} (\frac{3}{5}) = x, find the value of \cos x .$

Ans

$\begin{array}{l} Here, \sin^{- 1} \frac{3}{5} = x \\ \Rightarrow sinx = \frac{3}{5} \\ Now, \cos x = \sqrt{1 - \sin^{2} x} \\ = \sqrt{1 - {(\frac{3}{5})}^{2}} \\ = \sqrt{\frac{16}{25}} = \frac{4}{5} \end{array}$

Q.2

${Solvetan}^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4}$

Ans

$\begin{array}{l} \tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4} \\ \Rightarrow \tan^{- 1} [\frac{2 x + 3 x}{1 - 2 x \times 3 x}] = \frac{π}{4} [∴ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} [\frac{x + y}{1 - xy}]] \\ \Rightarrow \tan^{- 1} [\frac{5 x}{1 - 6 x^{2}}] = \frac{π}{4} \\ ∴ \frac{5 x}{1 - 6 x^{2}} = \tan \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \\ or 5 x = 1 - 6 x^{2} \\ \Rightarrow 6 x^{2} + 5 x - 1 = 0 \\ i . e . (6 x - 1) (x + 1) = 0 \\ Which gives, x = \frac{1}{6}, x = - 1 \\ Since x = -1 does not satisfy the equation as the L . H . S . of the equation \\ becomes negative, x = \frac{1}{6} is the only solution of the given equation . \end{array}$

Q.3

${Prove that tan}^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x + \sqrt{1 - x}}}) = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x$

Ans

$\begin{array}{l} Here, L . H . S . = {tam}^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) \\ Let x = \cos 2 θ \Rightarrow θ = \frac{1}{2} \cos^{- 1} x \\ Then L . H . S .= \tan^{- 1} [\frac{\sqrt{1 + \cos 2 θ} - \sqrt{1 - \cos 2 θ}}{\sqrt{1 + \cos 2 θ} + \sqrt{1 - \cos 2 θ}}] \\ = \tan^{- 1} [\frac{\sqrt{2 \cos^{2} θ} - \sqrt{2 \sin^{2} θ}}{\sqrt{2 \cos^{2} θ} + \sqrt{2 \sin^{2} θ}}] \\ = \tan^{- 1} [\frac{\sqrt{2} cosθ - \sqrt{2} sinθ}{\sqrt{2} cosθ + \sqrt{2} sinθ}] \\ = \tan^{- 1} [\frac{cosθ - sinθ}{cosθ + sinθ}] \\ = \tan^{- 1} [\frac{1 - tanθ}{1 + tanθ}] \\ = \tan^{- 1} [\frac{\tan \frac{π}{4} - tanθ}{1 + \tan \frac{π}{4} tanθ}] \\ = \tan^{- 1} [\tan^{- 1} (\frac{π}{4} - θ)] \\ = \frac{π}{4} - θ \\ = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x = R . H . S Proved \end{array}$

Q.4

${Prove that cot}^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] = \frac{x}{2}, \times \in (0, \frac{π}{4}) .$

Ans

$\begin{array}{l} L . H . S . \\ = \cot^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] \\ = \cot^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}} \times \frac{\sqrt{1 + sinx} + \sqrt{1 + sinx}}{\sqrt{1 + sinx} + \sqrt{1 - sinx}}] \\ = \cot^{- 1} [\frac{{(\sqrt{1 + sinx} + \sqrt{1 - sinx})}^{2}}{1 + sinx - 1 + sinx}] \\ = \cot^{- 1} [\frac{1 + sinx + 1 - sinx + 2 \sqrt{1 - \sin^{2} x}}{2 sinx}] \\ = \cot^{- 1} [\frac{2 (1 + \cos x)}{2 \sin x}] \\ = \cot^{- 1} [\frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}] \\ = \cot^{- 1} [\cot \frac{x}{2}] \\ = \frac{x}{2} = R . H . S \end{array}$

Q.5

${Show that sin}^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{63}{16} = π$

Ans

$\begin{array}{l} \sin^{- 1} \frac{12}{13} = x, \cos^{- 1} \frac{4}{5} = y \tan^{- 1} \frac{63}{16} = z \\ \Rightarrow \sin^{} x = \frac{12}{13}, \cos y = \frac{4}{5} and \tan z \frac{63}{16} \\ ∴ \cos x = \frac{5}{13}, \sin y = \frac{3}{4}, tanx = \frac{12}{5} and \tan y \frac{3}{4} \\ Now, \tan (x + y) = \frac{tanx + tany}{1 - \tan \times tany} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \times \frac{3}{4}} = \frac{- 63}{16} \\ \Rightarrow \tan (x + y) = - tanz \\ \Rightarrow \tan (x + y) = \tan (π - z) \\ ∴ x + y = π - z \\ \Rightarrow x + y + z = π \end{array}$

Q.6

$\begin{array}{l} Solve the following equation : \\ \tan^{- 1} (\frac{x - 1}{x - 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4} \end{array}$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{x - 1}{x - 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \times \frac{x + 1}{x + 2}}) = \frac{π}{4} \\ \Rightarrow \frac{x^{2} + x - 2 + x^{2} - x - 2}{x^{2} - 4 - x^{2} + 1} = 1 \\ \Rightarrow \frac{2 x^{2} - 4}{- 3} = 1 \\ \Rightarrow 2 x^{2} - 4 = - 3 \\ \Rightarrow x^{2} = \frac{1}{2} \\ \Rightarrow x = \pm \frac{1}{\sqrt{2}} \end{array}$

Q.7

$Evaluate \tan \frac{1}{2} [\sin^{- 1} \frac{2 x}{1 + x^{2}} + \cos^{- 1} \frac{1 - y^{2}}{1 + y^{2}}], |x| < 1, y > 0 and xy > 1$

Ans

$\begin{array}{l} Let x = \tan θ \\ \Rightarrow θ \tan^{- 1} x and = \tan^{- 1} y then, \\ \tan \frac{1}{2} [\sin^{- 1} \frac{2 x}{1 + x^{2}} + \cos^{- 1} \frac{1 - y^{2}}{1 + y^{2}}] \\ = \tan \frac{1}{2} [\sin^{- 1} (\frac{2 tanθ}{1 + \tan^{2} θ}) + \cos^{- 1} (\frac{1 - \tan^{2}}{1 + \tan^{2}})] \\ = \tan \frac{1}{2} [\sin^{- 1} (\sin 2 θ) + \cos^{- 1} (\cos 2)] \\ = \tan \frac{1}{2} (2 θ + 2) \\ = \tan (θ +) \end{array}$

Q.8

$Show that \sin^{- 1} \frac{3}{5} - \sin^{- 1} \frac{8}{17} = \cos^{- 1} \frac{84}{85}$

Ans

$\begin{array}{l} Let \sin^{- 1} \frac{3}{5} = x and \sin^{- 1} \frac{8}{17} = y \\ \Rightarrow \sin^{- 1} \frac{3}{5} = sinx and \frac{8}{17} = siny \\ as cosx = \sqrt{1 - \sin^{- 2} x} \\ = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \\ and \cos y = \sqrt{1 - \sin^{2} y} \\ = \sqrt{1 - \frac{64}{289}} = \frac{15}{17} \\ Now, \cos (x - y) = cosx cosy + \sin x siny \\ = \frac{4}{5} \times \frac{15}{17} + \frac{3}{5} \times \frac{8}{17} \\ \cos (x - y) = \frac{84}{85} \\ \Rightarrow x - y = \cos^{- 1} \frac{84}{85} \end{array}$

Q.9

$Evaluate \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2})$

Ans

$\begin{array}{l} \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2}) \\ \tan^{- 1} (\tan \frac{π}{4}) + \cos^{- 1} (- \cos \frac{π}{3}) + \sin^{- 1} (- \sin \frac{π}{6}) \\ = \frac{π}{4} + \cos^{- 1} (\cos (π - \frac{π}{3})) + \sin^{- 1} [\sin (\frac{- π}{6})] \\ we know range of \\ \tan^{- 1} \to (\frac{- π}{2}, \frac{π}{2}), \\ \cos^{- 1} \to [0, π] and \sin^{- 1} \to [\frac{- π}{2}, \frac{π}{2}] \\ ∴ \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2}) \\ = \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6} \\ = \frac{3 π + 8 π - 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4} \end{array}$

Q.10

$Write the function \tan^{- 1} (\frac{cosx - sinx}{cosx + sinx}) in the simplest form .$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{cosx - sinx}{cosx + sinx}) \\ = \tan^{- 1} (\frac{\frac{cosx}{cosx} - \frac{\sin x}{\cos x}}{\frac{cosx}{cosx} + \frac{sinx}{cosx}}) \\ = \tan^{- 1} (\frac{1 - tanx}{1 + tanx}) \\ = \tan^{- 1} (\frac{\tan \frac{π}{4} - tanx}{1 + \tan \frac{π}{4} tanx}) \\ = \tan^{- 1} (\tan (\frac{π}{4} - x)) [∴ \tan (A - B) = \frac{tanA - tanB}{1 + tanA tanB}] \\ = \frac{π}{4} - x \end{array}$

Q.11

${Express tan}^{-1} (\frac{cos x}{1-sin x}), \frac{π}{2} < x< \frac{π}{2} in the simplest form.$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{\cos x}{1 - \sin x}) = \tan^{- 1} [\frac{\sin (\frac{π}{2} - x)}{1 - \cos (\frac{π}{2} - x)}] \\ = \tan^{- 1} [\frac{\sin (\frac{π - 2 x}{2})}{1 - \cos (\frac{π - 2 x}{2})}] \\ = \tan^{- 1} [\frac{2 \sin (\frac{π - 2 x}{4})}{2 \sin^{2} (\frac{π - 2 x}{4})}] \\ = \tan^{- 1} [\frac{\cos (\frac{π - 2 x}{4})}{\sin (\frac{π - 2 x}{4})}] \\ = \tan^{- 1} [\cot (\frac{π - 2 x}{4})] \\ = \tan^{- 1} [\tan (\frac{π}{2} - [\frac{π - 2 x}{4}])] \\ = \tan^{- 1} [\tan (\frac{π}{4} + \frac{π}{2})] = \frac{π}{4} + \frac{x}{2} \end{array}$

Q.12

$Find the value of \sin^{- 1} (\sin \frac{2 π}{3})$

Ans

$\begin{array}{l} The range of inverse function is [\frac{- π}{2}, \frac{- π}{2}] . \\ Now, \sin^{- 1} [\sin (\frac{2 π}{3})] = \sin^{- 1} [\sin (π - \frac{π}{3})] \\ = \sin^{- 1} [\sin (\frac{π}{3})] \\ = \frac{π}{3} \end{array}$

Q.13

$Write \tan^{- 1} (\frac{1}{\sqrt{x^{2} - 1}}), |x| > 1 in the simplest form .$

Ans

$\begin{array}{l} Let x = cosec θ \\ \Rightarrow θ = {cosec}^{- 1} x \\ Then, \tan^{- 1} (\frac{1}{\sqrt{x^{2} - 1}}) = \tan^{- 1} (\frac{1}{\sqrt{{cosec}^{2} - 1}}) \\ = \tan^{- 1} (\frac{1}{cotθ}) \\ = \tan^{- 1} (tanθ) \\ = θ \\ = {cosec}^{- 1} x \end{array}$

Q.14

$Prove that \tan^{- 1} x + \tan^{- 1} \frac{2 x}{1 - x^{2}} = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$

Ans

$\begin{array}{l} Let x = \tan θ \\ Now, \\ L . H . S . = \tan^{- 1} x + \tan^{- 1} \frac{2 x}{1 - x^{2}} \\ = \tan^{- 1} (tanθ) + \tan^{- 1} (\frac{2 tanθ}{1 - \tan^{2} θ}) \\ = θ + \tan^{- 1} (\tan 2 θ) \\ = θ + 2 θ = 3 θ . \dots.. (1) \\ R . H . S .= \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (3 tanθ) = 3 θ . \dots. (2) \\ by (1) and (2) \\ L . H . S .= R . H . S . proved \end{array}$

Q.15

$Show that : \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{2}{11} = \tan^{- 1} \frac{3}{4}$

Ans

$\begin{array}{l} L . H . S .= \tan^{- 1} [\frac{\frac{1}{2} + \frac{2}{11}}{1 - \frac{1}{2} \times \frac{2}{11}}] [\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} \frac{15}{20} \\ = \tan^{- 1} \frac{3}{4} = R . H . S . \end{array}$

Q.16

$\begin{array}{l} Find the principal value of \\ (i) {sin}^{-1} (\frac{-1}{2}) \\ (ii) {cot}^{-1} (\frac{-1}{\sqrt{3}}) \end{array}$

Ans

$\begin{array}{l} (i) \sin^{- 1} (\frac{- 1}{2}) = y \\ \Rightarrow \sin y = \frac{- 1}{2} \\ \Rightarrow \sin y = - \sin \frac{π}{6} \\ \Rightarrow \sin y = - \sin (\frac{- π}{6}) or y = \frac{- π}{6} \\ We know that the range of inverse sine function is [\frac{- π}{2}, \frac{π}{2}] \\ ∴ \sin^{- 1} (\frac{- π}{6}) = \frac{- π}{6} \\ (ii) Let \cot^{- 1} (\frac{- 1}{\sqrt{3}}) = y \\ \Rightarrow coty = \frac{- 1}{\sqrt{3}} \\ \Rightarrow coty = \frac{- 1}{\sqrt{3}} \\ \Rightarrow coty = \cot (π - \frac{π}{3}) \\ \Rightarrow y = \frac{2 π}{3} \\ We know that range of inverse contangent function (0, π) \\ ∴ \cot^{- 1} (\frac{- 1}{\sqrt{3}}) = \frac{2 π}{3} \end{array}$

Q.17

$\begin{array}{l} Is the following statement true ? \\ \cos^{- 1} x = {(cosx)}^{- 1} \end{array}$

Ans

$No, \cos^{- 1} x ¹ \neq {(\cos x)}^{- 1} = \frac{1}{\cos x}$

Q.18

$Find the principal value of \cos^{- 1} (\frac{\sqrt{3}}{2})$

Ans

$\begin{array}{l} Let \cos^{- 1} (\frac{\sqrt{3}}{2}) = y \\ \Rightarrow \cos y = \frac{\sqrt{3}}{2} \\ Range of \cos^{- 1} is [0, π] \\ \cos \frac{π}{6} = \frac{\sqrt{3}}{2} \\ \Rightarrow \cos^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{6} \end{array}$

Q.19

$If \sin^{- 1} x = y, than what will be the range of y ?$

Ans

$\begin{array}{l} principal value of \sin^{- 1} lies between [- \frac{π}{2}, \frac{π}{2}] \\ - \frac{π}{2} \leq \frac{π}{2} y \leq \frac{π}{2} \end{array}$

Q.20

${Show that sin}^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \sin^{- 1} x$

Ans

$\begin{array}{l} \sin^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \sin^{- 1} x \\ let x = sinθ \\ \sin^{- 1} (2 x \sqrt{1 - x^{2}}) = \sin^{- 1} (2 sinθ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 sinθ cosθ) \\ = \sin^{- 1} (\sin 2 θ) \\ = 2 θ \\ = 2 \sin^{- 1} x \end{array}$

Q.21

$Find the value of \cos^{- 1} (\cos \frac{13 π}{6}) .$

Ans

$\begin{array}{l} \cos^{- 1} (\cos \frac{13 π}{6}) = \frac{13 π}{6} \\ But \frac{13 π}{6} \notin [0, π] \\ \cos (\frac{13 π}{6}) = \cos (2 π + \frac{π}{6}) = \cos (\frac{π}{6}) \\ ∴ \cos^{- 1} (\cos \frac{13 π}{6}) = \frac{π}{6} \end{array}$

Q.22

$Find the value of \tan^{- 1} (\frac{2}{11}) + \tan^{- 1} (\frac{7}{24}) .$

Ans

$\begin{array}{l} \tan^{- 1} (\frac{2}{11}) + \tan^{- 1} (\frac{7}{24}) . \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) [\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} (\frac{125}{250}) \\ = \tan^{- 1} (\frac{1}{2}) \end{array}$

Q.23

$Find the value of \tan^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}) .$

Ans

$\begin{array}{l} Let x = a \tan θ \\ Then \tan^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}) = \tan^{- 1} (\frac{3 a^{2} tanθ - a^{3} \tan^{3} θ}{a^{3} - 3 a^{2} \tan^{2} θ}) \\ = \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (\tan 3 θ) \\ =3 θ =3 \tan^{- 1} (\frac{x}{a}) \end{array}$

Q.24

$Write 2 \tan^{- 1} x in terms of \sin^{- 1} x, \cos^{- 1} x and \tan^{- 1} x .$

Ans

$2 \tan^{- 1} x = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$

Q.25

Write the domain and range of tan^-1x.

Ans

$\begin{array}{l} Domain of \tan^{- 1} x = R \\ Range of {ten}^{- 1} x = (- \frac{π}{2}, \frac{π}{2}) \end{array}$

Q.26

Fill in the following blanks:

${a)cos}^{- 1} (- x) = . \dots b) \cos^{- 1} \frac{1}{x} = . \dots.$

Ans

$\begin{array}{l} a) \cos^{- 1} (- x) = π - \cos^{- 1} x \\ b) \cos^{- 1} \frac{1}{x} = \sec^{- 1} x \end{array}$

Q.27

$Find the value of sin [\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})] .$

Ans

$\begin{array}{l} \sin [\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})] \\ = \sin [\frac{π}{3} + \sin^{- 1} (- \frac{1}{2})] \\ = \sin [\frac{π}{3} + \frac{π}{6}] \\ = \sin \frac{π}{2} \\ = 1 . \end{array}$

Q.28

$Evaluate \tan^{- 1} (\tan \frac{4 π}{3}) .$

Ans

$\begin{array}{l} The rage of inverse tangent function is (\frac{- π}{2}, \frac{π}{2}) . \\ Now, \tan^{- 1} [\tan (\frac{4 π}{3})] = \tan^{- 1} [\tan (π + \frac{π}{3})] \\ = \tan^{- 1} (\frac{π}{3}) \\ = \frac{π}{3} \end{array}$

Q.29

$Solve sin \{2 \cos^{- 1} (\cot (2 \tan^{- 1} x))\} = 0 .$

Ans

$\begin{array}{l} Given equation is \\ \sin \{2 \cos^{- 1} (\cot (2 \tan^{- 1} x))\} = 0 \\ \Rightarrow 2 \cos^{- 1} (\cot (2 \tan^{- 1} x)) = n^{π}, where n is any integer \\ \Rightarrow \cos^{- 1} (\cot (2 \tan^{- 1} x)) = n^{π / 2} \\ \Rightarrow \cos^{- 1} (\cot (2 \tan^{- 1} x)) = 0, \frac{π}{2}, π \\ (since \cos^{- 1} x lies in [0, π]) \\ \Rightarrow (\cot (2 \tan^{- 1} x)) = \cos 0, \cos \frac{π}{2}, \cos π = 1, 0, - 1 \\ \Rightarrow \cot [\tan^{- 1} \{(2 x) / (1 - x^{2})\}] = 1, 0, - 1 \\ \Rightarrow 1 / [\tan \{\tan^{- 1} [(2 x) / (1 - x^{2})]\}] = 1, 0, - 1 \\ \Rightarrow (1 - x^{2}) / (2 x) = 1, 0, - 1 \\ \Rightarrow (1 - x^{2}) / (2 x) = 1, (1 - x^{2}) / (2 x) = 0 and (1 - x^{2}) / (2 x) = - 1 \\ \Rightarrow x = - 1 \pm \sqrt{2}, x = \pm 1, x = 1 \pm \sqrt{2} . \end{array}$

Q.30

${Prove that sin}^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} = \sinh - 1 \frac{63}{65} .$

Ans

$\begin{array}{l} \sin^{- 1} \frac{12}{13} = x and \cos^{- 1} \frac{4}{5} = y \\ ∴ \cos x \frac{5}{13} and siny = \frac{3}{5} \\ Now, \sin (x + y) = sinx cosy + cosx siny \\ = \frac{12}{13}, \frac{4}{5} + \frac{5}{13}, \frac{3}{5} = \frac{63}{65} \\ \Rightarrow x + y = \sin^{- 1} \frac{63}{65} \\ so, \sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} = \sin^{- 1} \frac{63}{65} \end{array}$

Q.31

$Prove that \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} = \tan^{- 1} \frac{11}{29}$

Ans

$\begin{array}{l} Prove that \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} = \tan^{- 1} \frac{11}{29} \\ L . H . S .= \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} \\ = \tan^{- 1} [\frac{\frac{1}{5} + \frac{1}{6}}{1 - \frac{1}{5} \times \frac{1}{6}}] \\ = \tan^{- 1} [\frac{6 + 5}{30 - 1}] \\ = \tan^{- 1} [\frac{11}{29}] = R . H . S . Proved . \end{array}$

Q.32

${Find the principal value of sin}^{- 1} (\frac{1}{2}) .$

Ans

$\begin{array}{l} Let \sin^{- 1} (\frac{1}{2}) = θ \\ \Rightarrow \sin θ = \frac{1}{2} \\ \Rightarrow \sin θ = \sin (\frac{π}{6}) \\ \Rightarrow \sin θ = \sin (\frac{π}{6}) (âˆµ The range of \sin -^{- 1} x is [\frac{π}{2}, \frac{π}{2}]) \\ \Rightarrow θ = \frac{π}{6} \\ ∴ The principal value of \sin^{- 1} (\frac{1}{2}) is \frac{π}{6} . \end{array}$

Q.33

${Final the principal value of sin}^{- 1} (\frac{1}{\sqrt{2}}) .$

Ans

$\begin{array}{l} Let \sin^{- 1} (\frac{1}{\sqrt{2}}) = θ \\ \Rightarrow \sin θ = \frac{1}{\sqrt{2}} \\ \Rightarrow \cos θ = - \cos (\frac{π}{4}) \\ \Rightarrow \sin θ = \sin (\frac{π}{4}) (âˆµ The range of \sin -^{- 1} x is [\frac{π}{2}, \frac{π}{2}]) \\ \Rightarrow θ = \frac{π}{4} \\ ∴ The principal value of \sin^{- 1} (\frac{1}{\sqrt{2}}) is \frac{π}{4} . \end{array}$

Q.34

$Find the principal value of \cos^{- 1} (\frac{1}{\sqrt{2}}) .$

Ans

$\begin{array}{l} Let \cos^{- 1} (- \frac{1}{\sqrt{2}}) = 0 \\ \Rightarrow \cos θ = - \frac{1}{\sqrt{2}} \\ \Rightarrow \cos θ = - \cos (\frac{π}{4}) \\ \Rightarrow \cos θ = \cos (\frac{π}{4}) (âˆµ The range of \cos -^{- 1} x is [0, π]) \\ \Rightarrow \cos θ = \cos (\frac{3 π}{4}) \\ \Rightarrow θ = - \frac{3 π}{4} \\ ∴ The principal value of \cos^{- 1} (- \frac{1}{\sqrt{2}}) is \frac{3 π}{4} . \end{array}$

Q.35

$Evaluate the \cot (\tan^{- 1} a + \cot^{- 1} a) .$

Ans

$\begin{array}{l} We have, \\ = \cot (\tan^{- 1} a + \cot^{- 1} a) \\ = \cot (\frac{π}{2}) [âˆµ \tan^{- 1} a + \cot^{- 1} a = \frac{π}{2}] \\ = 0 \end{array}$

Q.36

$Evaluate the sin ({sin}^{-1} {x+cos}^{-1} x) .$

Ans

$\begin{array}{l} We have, \\ = \sin (\sin^{- 1} x + \cos^{- 1} x) \\ = \sin (\frac{π}{2}) [âˆµ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}] \\ = 1 \end{array}$

Q.37

$Put in the simplest form : \tan^{- 1} (\frac{1 - cosθ}{sinθ}) .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{1 - \cos θ}{\sin θ}) \\ = \tan^{- 1} (\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}) \\ = \tan^{- 1} (\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}) \\ = \tan^{- 1} (\tan \frac{θ}{2}) \\ = \frac{θ}{2} \end{array}$

Q.38

${Prove that the sin}^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}} .$

Ans

$\begin{array}{l} Let \sin^{- 1} x = θ . \dots.. (1) \\ \Rightarrow \sin θ = x \\ \Rightarrow \sin^{2} θ = x^{2} \\ \Rightarrow 1 - \cos^{2} θ = x^{2} \\ \Rightarrow \cos^{2} θ = 1 - x^{2} \\ \Rightarrow \cos θ = \sqrt{1 - x^{2}} \\ \Rightarrow \cos θ = \sqrt{1 - x^{2}} \\ \Rightarrow θ = \cos^{- 1} \sqrt{1 - x^{2}} . \dots. (2) \\ From (1) and (2) we get, \\ \sin^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}} \end{array}$

Q.39

${Prove that cos}^{- 1} (- x) = π - \cos^{- 1} (x) .$

Ans

$\begin{array}{l} Let \cos^{- 1} (- x) = θ . \dots. (1) \\ \Rightarrow - x = \cos θ \\ \Rightarrow x = - \cos θ \\ \Rightarrow x = - \cos (π - θ) \\ \Rightarrow π - θ = \cos^{- 1} x \\ \Rightarrow θ = π - \cos^{- 1} x . \dots (2) \\ From (1) and (2) we get, \\ \Rightarrow \cos^{- 1} (- x) = π - \cos^{- 1} x \end{array}$

Q.40

$Evaluate the cos ({cosec}^{- 1} x + \sec^{- 1} x) .$

Ans

$\begin{array}{l} We have, \\ \cos ({cosec}^{- 1} x + \sec^{- 1} x) \\ = \cos (\frac{π}{2}) [âˆµ {cosec}^{- 1} x + \sec^{- 1} x \frac{π}{2}] \\ = 0 \end{array}$

Q.41

${Evaluate the tan}^{- 1} 2 + \tan^{- 1} 3 .$

Ans

$\begin{array}{l} {We have, tan}^{- 1} 2 + {tan}^{- 1} 3 {=tan}^{- 1} (\frac{2 + 3}{1 - 2 \times 3}) \\ = {tan}^{- 1} (\frac{5}{1 - 6}) \\ = {tan}^{- 1} (\frac{5}{- 5}) \\ = {tan}^{- 1} (- 1) \\ = {tan}^{- 1} (1) \\ = \frac{π}{4} \end{array}$

Q.42

${Evaluate the tan}^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} .$

Ans

$\begin{array}{l} We have, \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{1 - \frac{14}{11 \times 24}}) \\ = \tan^{- 1} (\frac{125}{264 - 14}) \\ = \tan^{- 1} (\frac{125}{250}) \\ = \tan^{- 1} (\frac{1}{2}) \end{array}$

Q.43

$Prove that \sin^{- 1} x = {cosec}^{- 1} \frac{1}{x} .$

Ans

$\begin{array}{l} {Letsin}^{- 1} x = θ . \dots\dots. (1) \\ \Rightarrow \sin θ = x \\ \Rightarrow cosec θ = \frac{1}{x} \\ \Rightarrow θ = {cosec}^{- 1} = (\frac{1}{x}) . \dots\dots (2) \\ From (1) and (2) we get, \\ \sin^{- 1} x = {cosec}^{- 1} (\frac{1}{x}) \end{array}$

Q.44

$Put in the simplest form, \tan^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x} .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{\sqrt{1 + x^{2} - 1}}{x}) \\ Put x = tanθ \\ = \tan^{- 1} (\frac{\sqrt{1 + \tan^{2} θ - 1}}{tanθ}) \\ = \tan^{- 1} (\frac{\sqrt{\sec^{2} θ - 1}}{tanθ}) \\ = \tan^{- 1} (\frac{secθ - 1}{tanθ}) \\ = \tan^{- 1} (\frac{\frac{1 - cosθ}{cosθ}}{\frac{sinθ}{cosθ}}) \\ = \tan^{- 1} (\frac{1 - cosθ}{sinθ}) \\ = \tan^{- 1} (\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}) \\ = \tan^{- 1} (\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}) \\ = \tan^{- 1} (\tan \frac{θ}{2}) \\ = \frac{θ}{2} [âˆµ tanθ = x \Rightarrow θ = \tan^{- 1} x] \\ = \frac{1}{2} \tan^{- 1} x \end{array}$

Q.45

${Prove that sin}^{- 1} x + \cos^{- 1} x = \frac{π}{2} .$

Ans

$\begin{array}{l} We have to prove that \\ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} . \\ Let \sin^{- 1} x = θ . \dots.. (1) \\ \Rightarrow \sin θ = x \\ \Rightarrow \cos (\frac{π}{2} - θ) = x [âˆµ sinθ = \cos (\frac{π}{2} - θ)] \\ \Rightarrow \frac{π}{2} - θ = \cos^{- 1} x \\ \Rightarrow \cos^{- 1} x = \frac{π}{2} - θ \\ \Rightarrow θ + \cos^{- 1} x = \frac{π}{2} \\ \Rightarrow \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} (using (1)) \end{array}$

Q.46

$Show that \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{1}{3} = \frac{π}{4} .$

Ans

$\begin{array}{l} We have, \\ \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{1 - \frac{14}{11 \times 24}}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{125}{264 - 14}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{125}{250}) + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{3} \\ = \tan^{- 1} (\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}}) \\ = \tan^{- 1} (\frac{\frac{5}{6}}{\frac{5}{6}}) \\ = \tan^{- 1} (1) \\ = \frac{π}{4} \end{array}$

Q.47

$\begin{array}{l} If two angles of a triangle are \tan^{- 1} 2 and \tan^{- 1} 3, \\ than find the third angle . \end{array}$

Ans

$\begin{array}{l} Let third angle of a triangle be θ, then we have \\ θ + \tan^{- 1} + \tan^{- 1} 3 = π [âˆµ A + B + C = π] \\ \Rightarrow θ + \tan^{- 1} (\frac{2 + 3}{1 - 2 \times 3}) = π \\ \Rightarrow θ + \tan^{- 1} (\frac{5}{1 - 6}) = π \\ \Rightarrow θ + \tan^{- 1} (\frac{5}{- 5}) = π \\ \Rightarrow θ + \tan^{- 1} (1) = π \\ \Rightarrow θ + \frac{3 π}{4} = π [âˆµ \tan^{- 1} (- 1) = \frac{3 π}{4}] \\ \Rightarrow θ = π - \frac{3 π}{4} \\ \Rightarrow θ + \frac{π}{4} \\ Therefore, the third angle of a triangle is \frac{π}{4} . \end{array}$

Q.48

$Solve for x, \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{π}{2} .$

Ans

$\begin{array}{l} We have, \\ \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{π}{2} \\ \Rightarrow \sin^{- 1} (1 - x) = \frac{π}{2} + 2 \sin^{- 1} x \\ \Rightarrow (1 - x) = \sin (\frac{π}{2} + 2 \sin^{- 1} x) \\ \Rightarrow (1 - x) = \cos (2 \sin^{- 1} x) \\ \Rightarrow \cos^{- 1} (1 - x) = 2 \sin^{- 1} x \\ \Rightarrow \sin^{- 1} \sqrt{1 - {(1 - x)}^{2}} = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) \\ \Rightarrow \sqrt{1 - {(1 - x)}^{2}} = (2 x \sqrt{1 - x^{2}}) \\ On squaring both sides, \\ \Rightarrow 1 - {(1 - x)}^{2} = 4 x^{2} (1 - x^{2}) \\ \Rightarrow 1 - 1 - x^{2} + 2 x = 4 x^{2} - 4 x^{4} \\ \Rightarrow 4 x^{4} - 5 x^{2} + 2 x = 0 \\ \Rightarrow x (4 x^{3} - 5 x + 2) = 0 \\ \Rightarrow x = 0 or \\ 4 x^{3} - 5 x + 2 = 0 \\ Put x = . .. -1, 0, 1, 2, . \dots we get no real vlue of x \\ satisfy above cubic equation in [- 1, 1] \\ Hence x = 0 is only solution . \end{array}$

Q.49

$Solve for x, \tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4} .$

Ans

$\begin{array}{l} We have, \tan^{- 1} 2 x + \tan^{- 1} 2 x = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{2 x + 3 x}{1 - 2 x \times 3 x}) = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{5 x}{1 - 6 x^{2}}) = \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = \tan \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \\ \Rightarrow 5 x = 1 - 6 x^{2} \\ \Rightarrow 6 x^{2} = 5 x - 1 = 0 \\ \Rightarrow (6 x - 1) (x + 1) = 0 \\ \Rightarrow x = - (rejected as does not satisfy equation) \\ \Rightarrow x = \frac{1}{6} is the only solution of the given equation . \end{array}$

Q.50

$Put in the simplest form, \tan^{- 1} (\frac{\cos θ}{1 + \sin θ}) .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{\cos θ}{1 + \sin θ}) \\ = \tan^{- 1} (\frac{\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}}{{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}^{2}}) \\ = \tan^{- 1} (\frac{(\cos \frac{θ}{2} + \sin \frac{θ}{2}) (\cos \frac{θ}{2} - \sin \frac{θ}{2})}{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}) \\ = \tan^{- 1} (\frac{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})}) \\ Divide N^{1} and D^{1} by \cos \frac{θ}{2} \\ = \tan^{- 1} \frac{(1 - \tan \frac{θ}{2})}{(1 + \tan \frac{θ}{2})} \\ = \tan^{- 1} \frac{(\tan \frac{π}{4} - \tan \frac{θ}{2})}{(1 + \tan \frac{π}{4} . \tan \frac{θ}{2})} \\ = \tan^{- 1} [(\tan \frac{π}{4} - \frac{θ}{2})] \\ = \frac{π}{4} - \frac{θ}{2} \end{array}$

Q.51

$Show that \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8} = \frac{π}{4}$

Ans

$\begin{array}{l} We have, \\ \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8} \\ = \tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}) + \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}}) \\ = \tan^{- 1} (\frac{\frac{7 + 5}{35}}{1 - \frac{1}{35}}) + \tan^{- 1} (\frac{\frac{8 + 3}{24}}{1 - \frac{1}{24}}) \\ = \tan^{- 1} (\frac{\frac{12}{35}}{\frac{34}{35}}) + \tan^{- 1} (\frac{\frac{11}{24}}{\frac{23}{24}}) \\ = \tan^{- 1} (\frac{12}{34}) + \tan^{- 1} (\frac{11}{23}) \\ = \tan^{- 1} (\frac{6}{17}) + \tan^{- 1} (\frac{11}{23}) \\ = \tan^{- 1} (\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}}) \\ = \tan^{- 1} (\frac{\frac{138 + 187}{17 \times 23}}{\frac{391 - 66}{17 \times 23}}) \\ = \tan^{- 1} (\frac{325}{325}) \\ = \tan^{- 1} (1) \\ = \frac{π}{4} \\ = 1 \end{array}$

Q.52

$Put in the simplest form, \tan^{- 1} (\frac{\cos x}{1 - \sin x}) .$

Ans

$\begin{array}{l} We have, \tan^{- 1} (\frac{\cos x}{1 - \sin x}) \\ = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}) \\ \tan^{- 1} = (\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})}) \\ \tan^{- 1} = (\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})}) \\ Divide N^{1} and D^{1} by \cos \frac{x}{2} \\ = \tan^{- 1} \frac{(1 + \tan \frac{x}{2})}{(1 - \tan \frac{x}{2})} \\ = \tan^{- 1} \frac{(\tan \frac{π}{4} + \tan \frac{x}{2})}{(1 - \tan \frac{π}{4} . \tan \frac{x}{2})} \\ = \tan^{- 1} [(\tan \frac{π}{4} + \frac{x}{2})] \\ = \frac{π}{4} + \frac{x}{2} \end{array}$

Q.53

${Show that tan}^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{3}{4} = \tan^{- 1} 2 .$

Ans

$\begin{array}{l} LHS = \tan^{- 1} \frac{2}{11} + \tan^{- 1} \frac{7}{24} + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{1 - \frac{14}{11 \times 24}}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{125}{264 - 14}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{125}{250}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} \frac{3}{4} \\ = \tan^{- 1} (\frac{\frac{1}{2} + \frac{3}{4}}{1 - \frac{1}{2} \times \frac{3}{4}}) \\ = \tan^{- 1} (\frac{\frac{2 + 3}{4}}{\frac{8 - 3}{8}}) \\ = \tan^{- 1} (\frac{\frac{5}{4}}{\frac{5}{8}}) \\ = \tan^{- 1} (\frac{5}{4} \times \frac{8}{5}) \\ = \tan^{- 1} 2 (RHS Proved) \end{array}$

Q.54

$Evaluate \cot (\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8})$

Ans

$\begin{array}{l} We have, \\ \cot [\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}] \\ = \cot [\tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{8}] \\ = \cot [\tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}) + \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}})] \\ = \cot [\tan^{- 1} (\frac{\frac{7 + 5}{35}}{1 - \frac{1}{35}}) + \tan^{- 1} (\frac{\frac{8 + 3}{24}}{1 - \frac{1}{24}})] \\ = \cot [\tan^{- 1} (\frac{\frac{12}{35}}{\frac{34}{35}}) + \tan^{- 1} (\frac{\frac{11}{24}}{\frac{23}{24}})] \\ = \cot [\tan^{- 1} (\frac{12}{34}) + \tan^{- 1} (\frac{11}{23})] \\ = \cot [\tan^{- 1} (\frac{6}{17}) + \tan^{- 1} (\frac{11}{23})] \\ = \cot + [\tan^{- 1} (\frac{\frac{138 + 187}{17 \times 23}}{\frac{391 - 66}{17 \times 23}})] \\ = \cot + [\tan^{- 1} (\frac{325}{325})] \\ = \cot + [\tan^{- 1} (1)] \\ = \cot \frac{π}{4} \\ = 1 \end{array}$

Q.55

$\begin{array}{l} If \sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = \frac{π}{2}, then \\ find the value of x^{2} + y^{2} + z^{2} + 2 xyz . \end{array}$

Ans

$\begin{array}{l} We have, \\ \sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = \frac{π}{2} \\ \Rightarrow \cos^{- 1} \sqrt{1 - x^{2}} + \cos^{- 1} \sqrt{1 - y^{2}} = \frac{π}{2} \sin^{- 1} z \\ \Rightarrow \cos^{- 1} [\sqrt{1 - x^{2}} \sqrt{1 - y^{2}} - \sqrt{\{1 - (1 - x^{2})\}} \sqrt{\{1 - (1 - y^{2})\}}] = \cos^{- 1} z \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} - \sqrt{(x^{2})} \sqrt{(y^{2})} = z \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = xy + z \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = xy + z \\ Squaring both sides \\ {(xy + z)}^{2} = (1 - x^{2}) (1 - y^{2}) \\ \Rightarrow x^{2} y^{2} + z^{2} + 2 xyz = 1 - x^{2} - y^{2} + x^{2} y^{2} \\ \Rightarrow x^{2} + y^{2} + z^{2} + 2 xyz = 1 \\ Hence, value of x^{2} + y^{2} + z^{2} + 2 xyz is 1 . \end{array}$

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FAQs (Frequently Asked Questions)

1. Which chapter’s notes can I get for Class 12 mathematics on Extramarks?

Extramarks, an online learning platform, provides informative and detailed notes along with all topics and sub-topics of all the chapters for Mathematics Class 12. Students are advised to study with the help of these academic notes.

Chapter 1: Relations and Functions

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Applications of Derivatives

Chapter 7: Integrals

Chapter 8: Application of Integrals

Chapter 9: Differential Equations

Chapter 10: Vector Algebra

Chapter 11: Three Dimensional Geometry

Chapter 12: Linear programming

Chapter 13: Probability

2. Will using the Class 12 Mathematics Chapter 2 notes help me in my exam preparation?

Yes, definitely. The chapter 2 mathematics class 12 notes include every minute detail that the students must know in order to attain high scores. It includes solutions to all-important questions along with certain key points, formulae, derivations and more that will enhance the preparation level. Some tips, tricks and short-cut methods will also help students to tackle difficult questions in no time.

CBSE Class 12 Maths Revision Notes Chapter 2

Class 12 Mathematics Chapter 2 Notes

Key Topics Covered In Class 12 Mathematics Chapter 2 Notes

Inverse function:

Inverse trigonometric function:

Graphs of Inverse Trigonometric functions:

Properties:

Class 12 Mathematics Chapter 2 Notes Exercises & Answer Solutions.

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Key Features of Class 12 Mathematics Chapter 2 Notes

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CBSE Class 12 Maths Revision Notes Chapter 2

Class 12 Mathematics Chapter 2 Notes

Key Topics Covered In Class 12 Mathematics Chapter 2 Notes

Inverse function:

Inverse trigonometric function:

Graphs of Inverse Trigonometric functions:

Properties:

Class 12 Mathematics Chapter 2 Notes Exercises & Answer Solutions.

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics Chapter 2 Notes

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. Which chapter’s notes can I get for Class 12 mathematics on Extramarks?

2. Will using the Class 12 Mathematics Chapter 2 notes help me in my exam preparation?

Share

CBSE Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions