Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 7 Trigonometric Ratios Of Complementary Angles are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios Of Complementary Angles are extremely popular among Class 10 students for Math Trigonometric Ratios Of Complementary Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

Without using trigonometric tables, evaluate:
(i)
(ii)
(iii)
(iv)
(v)
(vi)

#### Question 2:

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 80° − sec 10° = 0
(iv) cosec272° − tan218° = 1
(v) cos275° + cos215° = 1
(vi) tan266° − cot224° = 0
(vii) sin248° + sin242° = 1
(viii) cos257° − sin233° = 0
(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0

#### Question 3:

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) sin35° sin55° − cos35° cos55° = 0
(v) (sin72° + cos18°)(sin72° − cos18°) = 0
(vi) tan48° tan23° tan42° tan67° = 1

Prove that:

#### Question 5:

Prove that:

(i)
(ii) $\frac{\mathrm{sin\theta }}{\mathrm{cos}\left(90°-\mathrm{\theta }\right)}+\frac{\mathrm{cos\theta }}{\mathrm{sin}\left(90°-\mathrm{\theta }\right)}=2$
(iii)
(iv) $\frac{\mathrm{cos}\left(90°-\mathrm{\theta }\right)\mathrm{sec}\left(90°-\mathrm{\theta }\right)\mathrm{tan\theta }}{\mathrm{cosec}\left(90°-\mathrm{\theta }\right)\mathrm{sin}\left(90°-\mathrm{\theta }\right)\mathrm{cot}\left(90°-\mathrm{\theta }\right)}+\frac{\mathrm{tan}\left(90°-\mathrm{\theta }\right)}{\mathrm{cot\theta }}=2$
(v) $\frac{\mathrm{cos}\left(90°-\mathrm{\theta }\right)}{1+\mathrm{sin}\left(90°-\mathrm{\theta }\right)}+\frac{1+\mathrm{sin}\left(90°-\mathrm{\theta }\right)}{\mathrm{cos}\left(90°-\mathrm{\theta }\right)}=2\mathrm{cosec\theta }$
(vi)
(vii)

#### Question 6:

Prove that:

(i) tan5° tan25° tan30° tan65° tan85° = $\frac{1}{\sqrt{3}}$
(ii) cot12° cot38° cot52° cot60° cot78° = $\frac{1}{\sqrt{3}}$
(iii) cos15° cos35° cosec55° cos60° cosec75° = $\frac{1}{2}$
(iv) cos1° cos2° cos3° ... cos180° = 0
(v) ${\left(\frac{\mathrm{sin}49°}{\mathrm{cos}41°}\right)}^{2}+{\left(\frac{\mathrm{cos}41°}{\mathrm{sin}49°}\right)}^{2}=2$

Disclaimer: The RHS of (v) given in textbook is incorrect. There should be 2 instead 1. The same has been corrected in the solution here.

#### Question 7:

Prove that
(i) sin (70° + θ) − cos (20° − θ) = 0
(ii) tan (55° − θ) − cot (35° + θ) = 0
(iii) cosec (67° + θ) − sec (23° − θ) = 0
(iv) cosec (65 °+ θ)  sec  (25° −  θ) − tan (55° − θ) + cot (35° + θ) = 0
(v) sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.

#### Question 8:

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

(i) sin67° + cos75°
(ii) cot65° + tan49°
(iii) sec78° + cosec56°
(iv) cosec54° + sin72°

#### Question 9:

If A, B and C are the angles of a $∆$ABC, prove that $\mathrm{tan}\left(\frac{\mathrm{C}+\mathrm{A}}{2}\right)=\mathrm{cot}\frac{\mathrm{B}}{2}$.

#### Question 10:

Hence, the value of $\theta$ is $15°.$

#### Question 11:

If sec2A = cosec(A $-$ 42$°$), where 2A is an acute angle, then find the value of A.       [CBSE2008]

We have,

Hence, the value of A is $44°$.

#### Question 12:

If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.

#### Question 13:

If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.

#### Question 14:

If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.

Prove that: