Without using trigonometrical tables, evaluate the following :
5 sin 50° sec 40°- 3 cos 59° cosec 31°

Question

Without using trigonometrical tables, evaluate the following : 5 sin 50° sec 40°- 3 cos 59° cosec 31°

KnowledgeBoat · Accepted Answer

Solving, ⇒ 5 sin 50° sec 40°- 3 cos 59° cosec 31° ⇒ 5 sin 50° sec (90° - 50°) - 3 cos 59° cosec (90° - 59°) [As, sec (90 - θ) = cosec θ and cosec (90 - θ) = sec θ] ⇒ 5 sin 50° cosec 50° - 3 cos 59° sec 59° ⇒ ⇒ 5 - 3 ⇒ 2. Hence, 5 sin 50° sec 40°- 3 cos 59° cosec 31° = 2.

Mathematics

Without using trigonometrical tables, evaluate the following :
5 sin 50° sec 40°- 3 cos 59° cosec 31°

Trigonometrical Ratios

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Mathematics

Without using trigonometrical tables, evaluate the following :5 sin 50° sec 40°- 3 cos 59° cosec 31°

Trigonometrical Ratios

Related Questions

Without using trigonometrical tables, evaluate the following :cos 18° sin 72° + sin 18° cos 72°

When 0° < A < 90°, solve the following equations :(i) sin 3A = cos 2A(ii) tan 5A = cot A

Without using trigonometrical tables, evaluate the following :
5 sin 50° sec 40°- 3 cos 59° cosec 31°

Without using trigonometrical tables, evaluate the following :
cos 18° sin 72° + sin 18° cos 72°

When 0° < A < 90°, solve the following equations :
(i) sin 3A = cos 2A
(ii) tan 5A = cot A