(i) sin 62° – cos 28°
(ii) cosec 35° – sec 55°.
Answer :
(i) sin 62° – cos 28°
= sin (90° – 28°) – cos 28°
= cos 28° – cos 28°
= 0
(ii) cosec 35° – sec 55°
= cosec 35° – sec (90° – 35°)
= cosec 35° – cosec 35°
= 0
More Solutions:
- cos 65°/sin 25° + cos 32°/sin 5
- Express each of the following in terms.
- sin2 28° – cos2 62° = 0
- sin 63° cos 27° + cos 63° sin 27° = 1
- sec 70° sin 20° – cos 20° cosec 70° = 0
- cot 54°/tan 36° + tan 20°/cot 70° – 2 = 0
- cos 80°/sin 10° + cos 59° cosec 31° = 2.
- Without using trigonometrical tables, evaluate:
- Prove the following: cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
- Simplify the following: It can be written as