RS Aggarwal 2021 2022 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 2021 2022 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 2021 2022 Solutions. All RS Aggarwal 2021 2022 Solutions for class 10 Maths are prepared by experts and are 100% accurate.

#### Question 1:

(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:

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

(iii) To Prove: cosec60° − sec30° = 0

(iv) To Prove: cot34° − tan56° = 0

(vi) To Prove: cos272° + cos218° = 1

#### Question 3:

(iii) To Prove: cosec60° − sec30° = 0

(iv) To Prove: cot34° − tan56° = 0

(vi) To Prove: cos272° + cos218° = 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:

(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

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

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

#### Question 5:

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

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

(i) sin72° + cosec72°

(ii) cosec66° + tan66°

(iii) tan68° + sec68°

(iv) cot59° + cosec59°

(v) cos51° + cot49° – sec47°

(vi) sin67° + cos75°

#### Question 6:

(i) sin72° + cosec72°

(ii) cosec66° + tan66°

(iii) tan68° + sec68°

(iv) cot59° + cosec59°

(v) cos51° + cot49° – sec47°

(vi) sin67° + cos75°

Given: sin3A = cos(A – 10°)

#### Question 7:

Given: sin3A = cos(A – 10°)

Given: tanA = cot(A + 10°)

#### Question 8:

Given: tanA = cot(A + 10°)

Given: cos2A = sin(A – 15°)

#### Question 9:

Given: cos2A = sin(A – 15°)

Given: sec4A = cosec(A – 15°)

#### Question 10:

Given: sec4A = cosec(A – 15°)

Given: tan2θ = cot(θ + 60°)

#### Question 11:

Given: tan2θ = cot(θ + 60°)

Given: sin(θ + 36°) = cosθ

#### Question 12:

Given: sin(θ + 36°) = cosθ

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#### Question 12:

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

#### Question 13:

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

#### Question 16:

Given: sinA = cosB

#### Question 17:

Given: sinA = cosB

Given: cos(α + β) = 0

#### Question 18:

Given: cos(α + β) = 0

#### Question 19:

Given: sec4A = cosec(– 10°)

#### Question 20:

Given: sec4A = cosec(– 10°)

Given: sin3A = cos(A – 10°)

#### Question 21:

Given: sin3A = cos(A – 10°)

(i) cot34° – tan56° = .......

(ii) cosec31° – sec59° = .......

(iii) cos267° + cos223° = ..........

(iv) cosec254° – tan236° = .........

(v) sec240° – cot250° = ..........

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