Trigonometrical Ratios of Complementary Angles: Problems and Solutions

🔸 01. If sin 23° = p, express sin 67° in terms of p.+

sin 23° = p

sin (90° − 67°) = p

cos 67° = p [∵ sin (90° − θ) = cos θ]

Again,

[∵ sin2θ + cos2θ = 1]

🔸 02. Find the value of cos220° + cos270°.+

cos220° + cos270°

= cos220° + cos2(90° − 20°)

= cos220° + sin220° [∵ cos (90° − θ) = sin θ]

= 1 [∵ sin2θ + cos2θ = 1]

🔸 03. Without using a logarithm table, find the value of tan 36° tan 54°.+

tan 36° tan 54°

= tan 36° tan (90° − 36°)

= tan 36° cot 36° [∵ tan (90° − θ) = cot θ]

= 1 [∵ tan θ cot θ = 1]

🔸 04. If ∠A + ∠B = 90°, prove that .+

L.H.S.,

= sin A [∵ sin θ cosec θ = 1]

= sin (90° − B) [∵ ∠A + ∠B = 90°]

= cos B [∵ sin (90° − B) = cos B]

= R.H.S.

🔸 05. Prove that cot (90° − A) cot A cos (90° − A) tan (90° − A) = cos A.+

L.H.S.,

cot (90° − A) cot A cos (90° − A) tan (90° − A)

= tan A cot A sin A cot A [∵ cot (90° − A) = tan A, cos (90° − A) = sin A and tan (90° − A) = cot A]

= 1 × sin A cot A [∵ tan θ cot θ = 1]

= [∵ ]

= cos A

= R.H.S.

🔸 06. Prove that tan 1° tan 2° tan 3° ... tan 87° tan 88° tan 89° = 1.+

L.H.S.,

tan 1° tan 2° tan 3° ... tan 87° tan 88° tan 89°

There are 89 factors in total, and the middle factor is tan 45°.

= tan 1° tan 2° tan 3° ... tan 44° tan 45° tan 46° ... tan 87° tan 88° tan 89°

= tan 1° tan 2° tan 3° ... tan 44° tan 45° tan (90° − 44°) ... tan (90° − 3°) tan (90° − 2°) tan (90° − 1°)

= tan 1° tan 2° tan 3° ... tan 44° tan 45° cot 44° ... cot 3° cot 2° cot 1° [∵ tan (90° − θ) = cot θ]

= 1 [∵ tan θ cot θ = 1 and tan 45° = 1]

= R.H.S.

🔸 07. If sin 4θ = cos 5θ and θ is an acute angle, find the value of θ.+

sin 4θ = cos 5θ

sin 4θ = sin (90° − 5θ) [∵ cos θ = sin (90° − θ)]

4θ = 90° − 5θ

4θ + 5θ = 90°

9θ = 90°

θ = 10°

🔸 08. Find the value of cos 1° cos 2° cos 3° ... cos 179°.+

cos 1° cos 2° cos 3° ... cos 179° contains the factor cos 90°.

cos 90° = 0

∴ If any factor of a product is zero, then the value of the whole product is zero.

Therefore, cos 1° cos 2° cos 3° ... cos 179° = 0.

🔸 09. Prove that cos273° − sin217° = 0.+

L.H.S.,

cos273° − sin217°

= cos273° − sin2(90° − 73°)

= cos273° − cos273° [∵ sin (90° − θ) = cos θ]

= 0

= R.H.S.

🔸 10. If sec θ = cosec φ and θ, φ are acute angles, find the value of sin (θ + φ).+

sec θ = cosec φ

sec θ = sec (90° − φ) [∵ cosec φ = sec (90° − φ)]

θ = 90° − φ

θ + φ = 90°

sin (θ + φ)

= sin 90°

= 1

Practice Problems

🔹 11. Find the value of sin222° + sin268° + cot230°.

🔹 12. If 2 cos (90° − A) − 1 = 0, show that sin 3A = 3 sin A − 4 sin3A.

🔹 13. Find the value of 3 cos 80° cosec 10° + 2 cos 59° cosec 31°.

🔹 14. Prove that tan 7° tan 23° tan 60° tan 67° tan 83° = √3.

🔹 15. Prove that .

🔹 16. Find the value of .

🔹 17. Find the value of .

🔹 18. Find the value of .

🔹 19. Find the value of .

🔹 20. Find the value of .

🔹 21. Find the value of .

🔹 22. In △ABC, prove that .

🔹 23. If ∠A + ∠B = 90°, prove that (cos A + cos B)2 = 1 + 2 cos A sin B.

🔹 24. If ∠A + ∠B = 90°, prove that .

🔹 25. If ∠A + ∠B = 90°, prove that .

🔹 26. If ∠A + ∠B = 90°, prove that .

🔹 27. If ∠A + ∠B = 90°, prove that sec2A + sec2B = sec2A sec2B.

🔹 28. Prove that cosec222° cot268° = sin222° + sin268° + cot268°.

🔹 29. If α + β = 90°, prove that .

🔹 30. If , prove that .

🔹 31. Prove that .

🔹 32. Prove that tan 1° tan 3° tan 5° ... tan 85° tan 87° tan 89° = 1.

🔹 33. Prove that .

🔹 34. If ∠A + ∠B = 90°, prove that .