Mathematics

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

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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°

⇒ $\text{5 sin 50°} \times \dfrac{1}{\text{sin 50°}} - \text{3 cos 59°} \times \dfrac{1}{\text{cos 59°}}$

⇒ 5 - 3

⇒ 2.

Hence, 5 sin 50° sec 40°- 3 cos 59° cosec 31° = 2.

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