Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 12 Trigonometric Ratios Of Some Complemantary Angles are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios Of Some Complemantary Angles are extremely popular among Class 10 students for Maths Trigonometric Ratios Of Some Complemantary Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Without using trigonometric tables, evaluate:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i) $\frac{\mathrm{sin}26°}{\mathrm{cos}64°}$

(iii) $\frac{\mathrm{tan}65°}{\mathrm{cot}25°}$

(iv) $\frac{\mathrm{cos}37°}{\mathrm{sin}53°}$

(vi) $\frac{\mathrm{cot}34°}{\mathrm{tan}56°}$

#### Question 2:

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 60° − sec 30° = 0
(iv) cot 34° − tan 56° = 0
(v) sin248° + sin242° = 1
(vi) cos272° + cos218° = 1

(iii) To Prove: cosec60° − sec30° = 0

(iv) To Prove: cot34° − tan56° = 0

(vi) To Prove: cos272° + cos218° = 1

#### Question 3:

Without using trigonometric tables, prove that:
(i) sin212° + sin2 78° = 1
(ii) sec229° – cot261° = 1
(iii) tan256° – cot234° = 0
(iv) cos257° – sin233° = 0
(v) sec250° – cot240° = 1
(vi) cosec272° – tan218° = 1

(i) To prove: sin212° + sin2 78° = 1

(ii) To prove: sec229° – cot261° = 1

(iii) To prove: tan256° – cot234° = 0

(iv) To prove: cos257° – sin233° = 0

(v) To prove: sec250° – cot240° = 1

(vi) To prove: cosec272° – tan218° = 1

#### Question 4:

Without using trigonometric tables, prove that:
(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2
(iv) tan 15° tan 60° tan 75° = $\sqrt{3}$
(v) tan48° tan23° tan42° tan67° tan 45° = 1
(vi) (sin72° + cos18°)(sin72° − cos18°) = 0
(vii) cosec 39° cos 51° + tan 21° cot 69° – sec221° = 0

(iv) To prove: tan15° tan60° tan75° = $\sqrt{3}$

(vii) To prove: cosec39° cos51° + tan21° cot69° – sec221° = 0

#### Question 5:

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.
(i) sin 72° + cosec 72°
(ii) cosec 66° + tan 66°
(iii) tan 68° + sec 68°
(iv) cot 59° + cosec 59°
(v) cos 51° + cot 49° – sec 47°
(vi) sin 67° + cos 75°

(i) sin72° + cosec72°

(ii) cosec66° + tan66°

(iii) tan68° + sec68°

(iv) cot59° + cosec59°

(v) cos51° + cot49° – sec47°

(vi) sin67° + cos75°

#### Question 6:

If sin 3A = cos (A – 10°), where 3A is an acute angle then find ∠A.

Given: sin3A = cos(A – 10°)

#### Question 7:

It tan A = cot (A + 10°), where (A + 10°) is acute then find ∠A.

Given: tanA = cot(A + 10°)

#### Question 8:

If cos 2A = sin (A – 15°), where 2A is acute then find ∠A.

Given: cos2A = sin(A – 15°)

#### Question 9:

If sec 4A = cosec (A – 15°), where 4A is acute then find ∠A.

Given: sec4A = cosec(A – 15°)

#### Question 10:

If tan 2θ = cot (θ + 60°), where 2θ and (θ + 6°) an acute angles, find the value of θ.

Given: tan2θ = cot(θ + 60°)

#### Question 11:

If sin (θ + 36°) = cos θ, where (θ + 36°) is acute, find θ.

Given: sin(θ + 36°) = cosθ

#### Question 12:

3cot 31° tan 15° cot 27° tan 75° cot 63° cot 59°

Ans

#### Question 19:

$\frac{\mathrm{cos}\left(90°-\mathrm{\theta }\right)\mathrm{sec}\left(90°-\mathrm{\theta }\right)\mathrm{tan\theta }}{\mathrm{cosec}\left(90°-\mathrm{\theta }\right)\mathrm{sin}\left(90°-\mathrm{\theta }\right)\mathrm{cot}\left(90°-\mathrm{\theta }\right)}+\frac{\mathrm{tan}\left(90°-\mathrm{\theta }\right)}{\mathrm{cot\theta }}$

Ans

#### Question 20:

sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

#### Question 1:

(a) $\frac{2}{3}$

(b) $\frac{16}{29}$

(c) $\frac{16}{3}$

(d) 1

#### Question 2:

(a) 0
(b) 1
(c) 2
(d) $\frac{3}{2}$

#### Question 3:

tan 10° tan 15° tan 75° tan 80° = ?
(a) $\sqrt{3}$
(b) 1
(c) $\frac{1}{\sqrt{3}}$
(d) –1

#### Question 4:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) 1
(b) $\sqrt{3}$
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{1}{2}$

#### Question 5:

cos 1° cos 2° cos 3° ... cos 180° = ?
(a) 0
(b) 1
(c) –1
(d) $\frac{1}{2}$

#### Question 6:

sin 43° cos 47° + cos 43° sin 47° = ?
(a) 0
(b) 1
(c) sin 4°
(d) cos 4°

#### Question 7:

sec 70° sin 20° + cos 20° cosec 70° = ?
(a) 0
(b) 1
(c) 2
(d) –2

#### Question 8:

cosec257° – tan233° = ?
(a) 1
(b) 0
(c) –1
(d) 2

#### Question 9:

sec210° – cot280° = ?
(a) 0

(b) 1

(c) $\frac{3}{4}$

(d) $\frac{1}{2}$

#### Question 10:

$\frac{2{\mathrm{sin}}^{2}63°+1+2{\mathrm{sin}}^{2}27°}{3{\mathrm{cos}}^{2}17°-2+3{\mathrm{cos}}^{2}73°}=?$

(a) $\frac{2}{3}$

(b) $\frac{3}{2}$

(c) 2

(d) 3

#### Question 11:

sin 38° – cos 52° = ?
(a) 0

(b) 1

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{2}{\sqrt{3}}$

#### Question 12:

(a) 3
(b) 2
(c) 1
(d) 0

Disclaimer: There must be $\mathrm{sin}31°$ instead of $\mathrm{sin}21°$.

#### Question 13:

(a) 2

(b) 1

(c) $\frac{2}{3}$

(d) $\frac{3}{2}$

#### Question 14:

(a) $\frac{1}{3}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $\frac{2}{\sqrt{3}}$

#### Question 15:

sin (60° + θ) – cos (30° – θ) = ?
(a) 2sin θ
(b) 2cos θ
(c) 0
(d) 1

#### Question 16:

If A and B are acute angles such that sin A = cos B then (A + B) = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Given: sinA = cosB

#### Question 17:

If cos (α + β) = 0 then sin (α – β) = ?
(a) sin 2α
(b) cos 2β
(c) sin α
(d) cos β

Given: cos(α + β) = 0

#### Question 18:

sin (45° + θ) – cos (45° – θ) = ?
(a) 0
(b) 1
(c) 2
(d) –2

#### Question 19:

If sec 4A = cosec (A – 10°) and 4A is acute then A = ?
(a) 20°
(b) 25°
(c) 30°
(d) 40°

Given: sec4A = cosec(– 10°)

#### Question 20:

If sin 3A = cos (A – 10°) and 3A is acute then A = ?
(a) 15°
(b) 20°
(c) 25°
(d) 30°

Given: sin3A = cos(A – 10°)

#### Question 21:

Fill in the blanks.
(i) cot 34° – tan 56° = .......
(ii) cosec 31° – sec 59° = .......
(iii) cos267° + cos223° = ..........
(iv) cosec254° – tan236° = .........
(v) sec240° – cot250° = ..........

(i) cot34° – tan56° = .......

(ii) cosec31° – sec59° = .......

(iii) cos267° + cos223° = ..........

(iv) cosec254° – tan236° = .........

(v) sec240° – cot250° = ..........

View NCERT Solutions for all chapters of Class 10