Important Questions Class 10 Mathematics Chapter 8

Important Questions Class 10 Mathematics Chapter 8 – Introduction to Trigonometry

Mathematics requires analytical thinking and problem-solving skills. One should do lots of practice and develop a deep understanding of the concepts of the subject in order to excel in Mathematics. As a result, students should have the right resources to refer to and follow accurate strategies. Students are advised to solve the  Important Questions Class 10 Mathematics Chapter 8 as it caters to varying levels of difficulty from easy to challenging questions covering all the topics from the chapter. 

Introduction to Trigonometry is a chapter that covers the basics of Trigonometry. The chapter covers the ratio of the sides of a right triangle with respect to its acute angle and the trigonometric ratios for angles of measure 0° and 90°. Chapter 8 Class 10 Mathematics important questions will cover questions  from the topics below: 

• Trigonometric Ratios
• Trigonometric Ratios of Some Specific Angles
• Trigonometric Ratios of Complementary Angles
• Trigonometric Identities

The important questions of Class 10 Mathematics Chapter 8 are made by Mathematics faculty experts who understand the key topics which students might find challenging while studying, therefore these questions are prepared in such a way that they can get to the point answers without wasting much time on a single subject. As a result, students can get a clear understanding of all the topics and improve their performance by solving these questions.

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Important Questions Class 10 Mathematics Chapter 8 – With Solutions

The following important questions and their solutions are included in the Mathematics Class 10 Chapter 8 important questions:

Question 1. when cos A = 4/5, the value for tan A is

(A) 3/5

(B) 3/4

(C) 5/3

(D) 4/3

Answer 1. (B) 3/4

Explanation: According to the question,

cos A = 4/5 …(1)

We know,

tan A = sinA/cosA

To find the value of sin A,

We have the equation,

sin2 θ +cos2 θ =1

So, sin θ = √ (1-cos2 θ)

Then,

sin A = √ (1-cos2 A) …(2)

sin2 A = 1-cos2 A

sin A = √(1-cos2 A)

Substituting equation (1) in (2),

We get,

Sin A = √(1-(4/5)2)

= √(1-(16/25))

= √(9/25)

= ¾

Therefore,

Tan A = (3/5)*(5/4) = (3/4)

 

Question 2. when sin A = ½, the value of cot A is

(A) √3

(B) 1

(C) √3/2

(D) 1/√3

Answer 2. (A) √3

Explanation: According to the question,

Sin A = ½ … (1)

We know that,

Cot A = 1/tan A = cos A/sin A …(2)

To find the value of cos A.

We have the equation,

sin2 θ +cos2 θ =1

So, cos θ = √(1-sin2 θ)

Then,

cos A = √(1-sin2 A) … (3)

cos2 A = 1-sin2 A

cos A = √ (1-sin2 A)

Substituting equation 1 in 3, we get,

cos A = √(1-1/4) = √(3/4) = √3/2

here, Substituting values of sin A and cos A in the equation 2, we get

cot A = (√3/2) × 2 = √3

 

Question 3. The value of the expression are as [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

(A) 1

(B) 0

(C) -1

(D) 3 2

Answer 3. (B) 0

Explanation: As per question,

we have to find out the value of the equation as,

cosec(75°+θ) – sec(15°-θ) – tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

hence, cosec (90°- θ) = sec θ

And, cot(90°-θ) = tan θ

We observe,

= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)

= 0

 

Question 4. Here,  sinθ = a b , so cosθ is equal to

(A) a/√(b2– a2)

(B) b/a

(C) √(b2-a2)/b

(D) b/√(b2-a2)

Answer 4.   (C) √(b2 – a2)/b

Explanation: According to the question,

sin θ =a/b

We know, sin2 θ +cos2 θ =1

sin2 A = 1-cos2 A

sin A = √(1-cos2 A

thus, cos θ = √(1-a2/b2 ) = √((b2–a2)/b2 ) = √(b2–a2 )/b

therefore, cos θ = √(b2 – a2 )/b

 

Question 5. If cos (α + β) = 0, then sin (α – β) can be reduced to

(A) cos β

(B) cos 2β

(C) sin α

(D) sin 2α

Answer 5.   (B) cos 2β

Explanation: According to the question,

cos(α+β) = 0

hence, cos 90° = 0

We can write,

cos(α+β)= cos 90°

By comparing the cosine equation on L.H.S and R.H.S,

We get,

(α+β)= 90°

α = 90°-β

Now we need to reduce sin (α -β ),

thus, we use,

sin(α-β) = sin(90°-β-β) = sin(90°-2β)

sin(90°-θ) = cos θ

thus, sin(90°-2β) = cos 2β

Therefore, sin(α-β) = cos 2β

 

Question 6. The value of the equation (tan1° tan2° tan3° … tan89°) is

(A) 0

(B) 1

(C) 2

(D) ½

Answer 6.   (B) 1

Explanation: tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

hence, tan 45° = 1,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

Since, tan(90°-θ) = cot θ,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

Since, tan θ = (1/cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44°). (1/tan 43°)… (1/tan 3°). (1/tan 2°). (1/tan 1°)

=(tan1° * 1/tan1°).(tan 2° * 1/tan 2°)…(tan 44° * 1/tan 44°)

= 1

thus, tan 1°.tan 2°.tan 3° …… tan 89° = 1

 

Question 7. If cos 9α = sinα and 9α < 90°, then the value of tan5α is

(A) 1/√3

(B) √3

(C) 1

(D) 0

Answer 7.   (C) 1

Explanation: According to the question,

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

We know that,

sin(90°-θ) = cos θ

So,

cos 9∝ = sin (90°-∝)

Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝

Thus, sin (90°-9∝) = sin∝

90°-9∝ =∝

10∝ = 90°

∝ = 9°

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5×9) = tan 45° = 1

∴, tan 5∝ = 1

 

Question 8. tan 47°/cot 43° = 1

Answer 8.   

True

Justification:

Since, tan (90° -θ) = cot θ

(tan 47°/tan 43°) = tan(90° – 47°)/cot 43°

(tan 47°/tan 43°) = (cot 43°/cot 43°) = 1

(tan 47°/cot 43°) = 1

 

Question 9. The value for the expression (cos223° – sin267°) will be positive.

Answer 9.   

False

Justification:

Since, (a2-b2) = (a+b)(a-b)

cos2 23° – sin2 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

 

Question 10. The value for the expression (sin 80° – cos 80°) is negative.

Answer 10.   

False

clarification:

We know that,

sin θ increases if 0° ≤ θ ≤ 90°

cos θ decreases if 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value – decreasing value)

= a positive value.

Thus, (sin 80°-cos 80°) > 0.

 

Question 11. If cosA + cos2A = 1, then sin2A + sin4A = 1.

Answer 11.   

True

Justification:

According to the question,

cos A+cos2 A = 1

i.e., cos A = 1- cos2 A

Since,

sin2 θ+cos2 θ = 1

sin2 θ = 1- cos2 θ)

We get,

cos A = sin2 A …(1)

Squaring L.H.S and R.H.S,

cos2 A = sin4 A …(2)

To find sin2A+sin4 A=1

Adding equations (1) and (2),

We get

sin2A + sin4 A= cos A + cos2 A

Therefore, sin2A+ sin4 A = 1

 

Question 12. (tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec2 θ.

Answer 12.   

False

Justification:

L.H.S = (tan θ+2) (2 tan θ+1)

= 2 tan2 θ + tan θ + 4 tan θ + 2

= 2 tan2θ+5 tan θ+2

Since, sec2 θ – tan2 θ = 1, we get, tan2θ = sec2 θ-1

= 2(sec2 θ-1) +5 tan θ+2

= 2 sec2 θ-2+5 tan θ+2

= 5 tan θ+ 2 sec2 θ ≠R.H.S

∴, L.H.S ≠ R.H.S

 

Question 13. (√3+1) (3 – cot 30°) = tan3 60° – 2 sin 60°

Answer 13.   

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]

= ((√3)2– 1) √3 [∵ (√3+1)(√3-1) = ((√3)2 – 1)]

= (3-1) √3

= 2√3

Similarly solving R.H.S: tan3 60° – 2 sin 60°

Since, tan 60o = √3 and sin 60o = √3/2,

We get,

(√3)3 – 2.(√3/2) = 3√3 – √3

= 2√3

Therefore, L.H.S = R.H.S

Hence, proved.

 

Question 14. If 1 + sin2θ = 3sinθ cosθ , then prove that tanθ = 1 or ½.

Answer 14.   

Given: 1+sin2 θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin2 θ,

We get,

(1+sin2 θ)/sin2 θ  = 3 sin θ cos θ/ sin2 θ

⇒ (1/sin2 θ) + 1 = 3cos θ/ sin θ

cosec2 θ + 1 = 3 cot θ

Since,

cosec2 θ – cot2 θ = 1 ⇒ cosec2 θ = cot2 θ +1

⇒ cot2 θ +1+1 = 3 cot θ

⇒ cot2 θ +2 = 3 cot θ

⇒ cot2 θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot2 θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ – 1)(cot θ – 2) = 0

⇒ cot θ = 1, 2

Since,

tan θ = 1/cot θ

tan θ = 1, ½

Hence, proved.

 

Question 15. Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Answer 15.   

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)2 = 1

⇒ sin2 θ + 4 cos2 θ + 4sin θcos θ = 1

Since, sin2 θ = 1 – cos2 θ and cos2 θ = 1 – sin2 θ

⇒ (1 – cos2 θ) + 4(1 – sin2 θ) + 4sin θcos θ = 1

⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4sin θcos θ = 1

⇒ – 4 sin2 θ – cos2 θ + 4sin θcos θ = – 4

⇒ 4 sin2 θ + cos2 θ – 4sin θcos θ = 4

We know that,

a2+ b2 – 2ab = ( a – b)2

So, we get,

(2sin θ – cos θ)2 = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

 

Question 16. In ∆ ABC, the right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

Answer 16.   

In the given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm as well as BC = 7 cm

According to the Pythagoras Theorem,

In the right-angled triangle, the squares of the hypotenuse sides are equal to the addition of the squares for the other two sides.

on applying the Pythagoras theorem, we observe

AC2=Ab2+BC2

AC2 = (24)2+72

AC2 = (576+49)

AC2 = 625cm2

AC = √625 = 25

thus, AC = 25 cm

(i) To find Sin (A), Cos (A)

We get that sine (or) Sin function is equal to the ratio of length of the opposite sides to the hypotenuse sides. So it becomes

Sin (A) = Opposite side /Hypotenuse side = BC/AC = 7/25

Cosine or Cos function is same as the ratio of the length of the adjacent side to the hypotenuse side so it becomes,

Cos (A) = Adjacent side/Hypotenuse side = AB/AC = 24/25

(ii) To find out Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

 

Question 17. In Fig. find tan P – cot R

Answer 17.   

In the given triangle QPR, the given triangle is right angled at Q, and the given measures are:

PR = 13cm,

PQ = 12cm

Since the given triangle is the right-angled triangle, to find the side QR, apply the Pythagorean theorem

According to the Pythagorean theorem,

In the right-angled triangle, the squares of the hypotenuse side is equal to the addition of the squares for the other two sides.

PR2 = QR2 + PQ2

Substitute the values for PR and PQ

132 = QR2+122

169 = QR2+144

hence, QR2 = 169−144

QR2 = 25

QR = √25 = 5

hence, the side QR = 5 cm

To find tan P – cot R:

According to trigonometric ratio, the tangent function is same as the ratio of the length of the opposite sides to the adjacent sides, the value for tan (P) becomes

tan (P) = Opposite sides /Adjacent sides = QR/PQ = 5/12

hence, cot function is the reciprocal of the tan function, then the ratio of the cot function becomes,

Cot (R) = Adjacent sides/Opposite sides = QR/PQ = 5/12

hence,

tan (P) – cot (R) = 5/12 – 5/12 = 0

hence, tan(P) – cot(R) = 0

 

Question 18. If sin A = 3/4, so Calculate cos A and tan A.

Answer 18.   

Let us assume that a right-angled triangle ABC, right angled at B

Given: Sin A = 3/4

We get that, Sin function is equal to the ratio of length of the opposite side to the hypotenuse side.

hence, Sin A = Opposite side /Hypotenuse side = 3/4

Let’s BC be 3k, and AC will be 4k

where k is a positive real number.

According to Pythagoras theorem, the squares of the hypotenuse side is same as the sum of the squares for the other two sides of the right angle triangle, and we observe,

AC2=Ab2 + BC2

Substitute the value of AC and BC

(4k)2=Ab2 + (3k)2

16k2−9k2 =Ab2

Ab2=7k2

hence, AB = √7k

then, we have to find the value of cos A and tan A

We get that,

Cos (A) = Adjacent side/Hypotenuse side

here, Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we observe

AB/AC = √7k/4k = √7/4

hence, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the value of the line BC and AB and cancel the constant k in both numerator and denominator, we observe,

BC/AB = 3k/√7k = 3/√7

hence, tan A = 3/√7

 

Question 19. Given 15 cot A = 8, find out sin A and sec A.

Answer 19.   

Let us assume a right-angled triangle ABC, right angled at B

Given: 15 cot A = 8

So, Cot A = 8/15

We get that cot function is equal to the ratio of length of the adjacent side to the opposite side.

Hence, cot A = Adjacent sides /Opposite sides = AB/BC = 8/15

Let AB be 8k as well as BC will be 15k

Here, k is the positive real number.

According to the Pythagoras theorem, so the squares of the hypotenuse side is same as the addition of the squares for the other two sides of the right angle triangle, and we observe,

AC2=Ab2 + BC2

Substitute the value of AB and BC

AC2= (8k)2 + (15k)2

AC2= 64k2 + 225k2

AC2= 289k2

Hence, AC = 17k

Then, we have to find the value of sin A and sec A

We get that,

Sin (A) = Opposite side /Hypotenuse

Here, Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we observe

Sin A = BC/AC = 15k/17k = 15/17

Hence, sin A = 15/17

Hence, secant or sec function is the reciprocal of the cost function which is the same as the ratio for the length of the hypotenuse side to adjacent side.

Sec (A) = Hypotenuse/Adjacent side

here, Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we observe,

AC/AB = 17k/8k = 17/8

hence, sec (A) = 17/8

 

Question 20. Given sec θ = 13/12 Calculate all the other trigonometric ratios

Answer 20.   

We get that sec function is the reciprocal of the cost function which is same as the ratio of the length of hypotenuse sides to the adjacent sides

Let us assume a right-angled triangle ABC, right angled at B

sec θ =13/12 = Hypotenuse side/Adjacent side = AC/AB

Let’s AC be 13k, and AB will be 12k

here, k is a positive real number.

According to Pythagoras theorem, the squares of the hypotenuse side is equal to the addition of the squares for the other two sides of the right angle triangle and we observe,

AC2=Ab2 + BC2

here, Substitute the value of AB and AC

(13k)2= (12k)2 + BC2

169k2= 144k2 + BC2

169k2= 144k2 + BC2

BC2 = 169k2 – 144k2

BC2= 25k2

hence, BC = 5k

then, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = Opposite Side/Hypotenuse Side = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse Side = AB/AC = 12/13

tan θ = Opposite Sides/Adjacent Sides = BC/AB = 5/12

Cosec θ = Hypotenuse Side/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Sides /Opposite Sides = AB/BC = 12/5

 

Question 21. If ∠A and ∠B are the acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Answer 21.   

Let us assume that the triangle ABC in which CD⊥AB

Given that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cost ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

then consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying the Pythagoras theorem in △CAD and △CBD we get,

CD2 = BC2 – BD2 … (3)

CD2 =AC2 −AD2 ….(4)

From the equations (3) and (4) we observe,

AC2−AD2 = BC2−BD2

then substitute the equations (1) and (2) in (3) and (4)

k2(BC2−BD2)=(BC2−BD2) k2=1

Putting this value in equation, we obtain that

AC = BC

∠A=∠B (Angles opposite to the equal side are equal-isosceles triangle)

 

Question 22. In triangle ABC, right-angled at B, when tan A = 1/√3 find out the value :

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Answer 22.   

Let’s ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let’s BC = 1k and AB = √3 k,

here k is the positive real number of the problem

By the Pythagoras theorem in ΔABC we get:

AC2=Ab2+BC2

AC2=(√3 k)2+(k)2

AC2=3k2+k2

AC2=4k2

AC = 2k

then find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

now, find the values of cos C and sin C

Sin C = AB/AC = √3/2

Cos C = BC/AC = 1/2

then, substitute the values in the given problems

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0

 

Question 23. In ∆ QPR, right-angled at Q, PR + QR = 25 cm as well as PQ = 5 cm. Determine the value of sin P, cos P and tan P

Answer 23.   

In the given triangle PQR, right angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

then let us assume, QR = x

PR = 25-QR

PR = 25- x

hence, According to the Pythagorean Theorem,

PR2 = PQ2 + QR2

Substitute the value of PR as x

(25- x)2 = 52 + x2

252 + x2 – 50x = 25 + x2

625 + x2-50x -25 – x2 = 0

-50x = -600

x= -600/-50

x = 12 = QR

then, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

then, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse Side = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse Side = PQ/PR = 5/13

(3) Tan p =Opposite Sides/Adjacent sides = QR/PQ = 12/5

 

Question 24. State whether the following will be true or false. clarify your answer.

(i) The value for tan A is always less than 1.

(ii) sec A = 12/5 thus some value of the angle A.

(iii)cos A is the abbreviation used to the cosecant of the angle A.

(iv) cot A is a product of cot A.

(v) sin θ = 4/3 for some of the angles  θ.

Answer 24.   

(i) The value of tan A will always be less than 1.

Answer: False

Proof: In ΔMNC the angle ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 that is greater than.

The triangle can be formed with the sides equal to 3, 4 and hypotenuse = 5 according to the Pythagoras theorem.

MC2=MN2+NC2

52=32+42

25=9+16

25 = 25

(ii) sec A = 12/5 thus, some value of the angle A

Answer: True

clarification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, here k is a positive real number.

By the Pythagoras theorem we observe,

MC2=MN2+NC2

(12k)2=(5k)2+NC2

NC2+25k2=144k2

NC2=119k2

Such a triangle is possible as it would follow the Pythagoras theorem.

(iii) cos A is an abbreviation used for the cosecant of the angle A.

Answer: False

clarification: Abbreviation used for the cosecant of angle M is cosec M. cos M is the abbreviation used for the cosine of angle M.

(iv) cot A is the product of the cot and A.

Answer: False

clarification: cot M is not the product of cot and M. It is the cotangent of ∠M.

(v) sin θ = 4/3 for some angle θ.

Answer: False

clarification: sin θ = Opposite/Hypotenuse

We know that in the right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always be less than 1 and it can never be 4/3 for any value of θ.

 

Question 25. Choose the correct option and clarify  your answer: 

(i) 2tan 30°/1+tan230° =

(A) sin 60°            (B) sin 45°          (C) tan 60°            (D) cos 30°

(ii) 1-tan245°/1+tan245° =

(A) tan 90°            (B) 0                    (C) sin 45°            (D) 1

(iii)  sin 2A = 2 sin A will true if A =

(A) 0°                   (B) 30°                  (C) 45°                 (D) 60°

(iv) 2tan30°/1-tan230° =

(A) cos 60°          (B) sin 45°             (C) tan 60°           (D) sin 30°

Answer 25.   

(i) (A) is correct.

here, Substitute for tan 30° in a given equations

tan 30° = 1/√3

2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2

= (2/√3)/(1+1/3) = (2/√3)/(4/3)

= 6/4√3 = √3/2 = sin 60°

we obtained that the solution is equivalent to the trigonometric ratio as sin 60°

(ii) (D) is correct.

Substitute the of tan 45° in given equation

tan 45° = 1

1-tan245°/1+tan245° = (1-12)/(1+12)

= 0/2 = 0

The solution of the above equations is 0.

(iii) (A) is correct.

To find out the value for A, substitute the degree given in the option one by one

sin 2A = 2 sin A is true if A = 0°

As sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

or,

Apply in the sin 2A formula, to find the degree value

sin 2A = 2sin A cos A

⇒2sin A cos A = 2 sin A

⇒ 2cos A = 2 ⇒ cos A = 1

then, we have to check, to get the solution as 1, which degree value has to be applied.

When 0 degree is applied to the cos value, i.e., cos 0 =1

Therefore, ⇒ A = 0°

(iv) (C) is correct.

Substitute the of tan 30° in given equations

tan 30° = 1/√3

2tan30°/1-tan230° =  2(1/√3)/1-(1/√3)2

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

The value for the given equation will be equivalent to tan 60°.

 

Question 26. when tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

Answer 26.   

tan (A + B) = √3

Since √3 = tan 60°

then, substitute the degree values

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equations (i)

tan (A – B) = 1/√3

hence, 1/√3 = tan 30°

Now substitute the degree values

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

then add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

now, substitute the value for A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

 

Question 27. State whether the following will be true or false. clarify your answers.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ will increase.

(iii) The value of cos θ increases as θ will increase.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Answer 27. (i) False.

clarification:

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we observe

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

Hence the values obtained are not equal, the solution is false.

(ii) True.

clarification:

As per the values obtained as per the unit circle, the values of sin will:

sin 0° = 0

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Therefore, the value of sin θ increases as θ increases. thus, the statement is true

(iii) False.

As per the values obtained as per the unit circle, the values of cos will:

cos 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Therefore, the value of cos θ decreases as θ will increase. So, the statement given above is false.

(iv) False

sin θ = cos θ, if a right triangle has 2 angles of (π/4). Thus, the above statement is false.

(v) True.

hence cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A/sin A

then substitute A = 0°

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Thus, it is true

 

Question 28. Evaluate :

(i) sin 18°/cos 72°

(ii) tan 26°/cot 64°

(iii)  cos 48° – sin 42°

(iv)  cosec 31° – sec 59°

Answer 28. (i) sin 18°/cos 72°

To simplify it, convert the sin function into the cos function

We get that, 18° is written as 90° – 18°, which is same as the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify the equation

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

To simplify it, convert the tan function into cot function

We get that, 26° is written as 90° – 26°, which is equal to the cot 64°.

= tan (90° – 26°)/cot 64°

Substitute the value, to simplify the equation

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

To simplify it, convert the cos function into the sin function

We get that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

Substitute the value, to simplify the equation

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify it, convert the cosec function into the sec function

We get that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

Substitute the value, to simplify the equation

= sec 59° – sec 59° = 0

 

Question 29. Shown in the equation:

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Answer 29. (i) tan 48° tan 23° tan 42° tan 67°

here, we Simplify the given problem by converting some of the tan functions into the cot functions

We get that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute to the value

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

 

(ii) cos 38° cos 52° – sin 38° sin 52°

here, Simplify the given problem by converting some of the cos functions to the sin functions

We get,

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

here, Substitute the value

= sin 52° sin 38° – sin 38° sin 52° = 0

 

Question 30. when tan 2A = cot (A – 18°), where 2A is the acute angle, find the value of A.

Answer 30.   tan 2A = cot (A- 18°)

We get that tan 2A = cot (90° – 2A)

Substitute the above equation in given problem

⇒ cot (90° – 2A) = cot (A -18°)

then, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

hence, the value of A = 36°

 

Question 31. If tan A = cot B, proved that A + B = 90°.

Answer 31.   tan A = cot B

We get that cot B = tan (90° – B)

To proved A + B = 90°, substitute the above equation in given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

thus Proved.

 

Question 32. when sec 4A = cosec (A – 20°), here 4A is an acute angle, findout the value of A.

Answer 32.   sec 4A = cosec (A – 20°)

We get that sec 4A = cosec (90° – 4A)

To find out the value of A, thus, substitute the above equation in a given problems

cosec (90° – 4A) = cosec (A – 20°)

then, equate the angles

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

hence, the value of A = 22°

 

Question 33. when A, B and C are interior angles of the triangle ABC, then show that

    sin (B+C/2) = cos A/2

Answer 33. We get that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find out the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

then, multiply both sides by the sin functions, we observe

⇒ sin (B+C)/2 = sin (90°-A/2)

hence,  sin (90°-A/2) = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

therefore, proved.

 

Question 34. Express sin 67° + cos 75° in the terms of the trigonometric ratios of angles in between 0° and 45°.

Answer 34.   Given:

sin 67° + cos 75°

In term of sin as the cos function and cos as sin function, it can be written as  

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

then , simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is expressed as cos 23° + sin 15°

 

Question 35. Express the trigonometric ratios of sin A, sec A and tan A in terms of cot A.

Answer 35.   F To  convert the given trigonometric ratios in terms of cost functions, use trigonometric formulas

We get that,

cosec2A – cot2A = 1

cosec2A = 1 + cot2A

hence, cosec function is the inverse of sin function, it is written as

1/sin2A = 1 + cot2A

then, rearrange the terms, it becomes

sin2A = 1/(1+cot2A)

then, take square roots on both sides, we get

sin A = ±1/(√(1+cot2A)

The above equation define that the sin function in terms of cot function

then, to express sec function in terms of cot function, use this formula

sin2A = 1/ (1+cot2A)

then, represent the sin function as cos function

1 – cos2A = 1/ (1+cot2A)

Rearrange the terms,

cos2A = 1 – 1/(1+cot2A)

⇒cos2A = (1-1+cot2A)/(1+cot2A)

hence, sec function is the inverse of cos function,

⇒ 1/sec2A = cot2A/(1+cot2A)

Take the reciprocal and the square roots on both sides, we get

⇒ sec A = ±√ (1+cot2A)/cotA

then, to express the tan function in the terms of the cot function

tan A = sin A/cos A as we cot A = cos A/sin A

hence, cot function is the inverse of tan function, thus, it is rewrite are as

tan A = 1/cot A

 

Question 36. Write the other trigonometric ratios of ∠A in the terms of sec A.

Answer 36.   Cos A function in the terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in the terms of sec A:

cos2A + sin2A = 1

here, Rearrange the terms

sin2A = 1 – cos2A

sin2A = 1 – (1/sec2A)

sin2A = (sec2A-1)/sec2A

sin A = ± √(sec2A-1)/sec A

cosec A function in the terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec2A-1)

then, tan A function in terms of sec A:

sec2A – tan2A = 1

Rearrange the terms

⇒ tan2A = sec2A – 1

tan A = √(sec2A – 1)

cot A function in the terms for sec A:

tan A = 1/cot A

⇒ cot A = 1/tan A

cot A = ±1/√(sec2A – 1)

 

Question 37. Evaluate that:

(i) (sin263° + sin227°)/(cos217° + cos273°)

(ii)  sin 25° cos 65° + cos 25° sin 65°

Answer 37.   

(i) (sin263° + sin227°)/(cos217° + cos273°)

To simplify it, convert some of the sin functions into the cos functions and cos function into the sin function and it become as,

= [sin2(90°-27°) + sin227°] / [cos2(90°-73°) + cos273°)]

= (cos227° + sin227°)/(sin227° + cos273°)

= 1/1 =1                       (hence sin2A + cos2A = 1)

hence, (sin263° + sin227°)/(cos217° + cos273°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

for simplify it, convert some of the sin functions into the cos functions and the cos function into sin function and it become as,

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos265° + sin265° = 1 (since sin2A + cos2A = 1)

hence, sin 25° cos 65° + cos 25° sin 65° = 1

 

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