Solve for 'θ': sec ( θ 2 + 10 ° ) = 2 √ 3

Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`

Sum

`sec(θ/2 + 10°) = (2)/sqrt(3)`

⇒ `sec(θ/2 + 10°)` = sec 30°

⇒ `θ/(2) + 10°` = 30°

⇒ `θ/(2)` = 20°
⇒ θ = 40°.

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Trigonometric Equation Problem and Solution

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Chapter 22: Trigonometrical Ratios of Standard Angles - Exercise 27.1

Chapter 22 Trigonometrical Ratios of Standard Angles
Exercise 27.1 | Q 7.3

If sin 3A = 1 and 0 < A < 90^°, find cos 2A

Solve for x : sin (x + 10°) = `(1)/(2)`

Solve for x : cos² 30° + sin² 2x = 1

Find the value of 'A', if cosec 3A = `(2)/sqrt(3)`

If θ = 30°, verify that: sin 3θ = 4sinθ . sin(60° - θ) sin(60° + θ)

In right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if:
a. AB is `sqrt(3)` times of BC.
B. BC is `sqrt(3)` times of BC.

Evaluate the following: sin28° sec62° + tan49° tan41°

Evaluate the following: sin22° cos44° - sin46° cos68°

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cos72° - cos88°

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cos84° + cosec69° - cot68°