Question

Find value of
$\sec ^{ - 1} (2 \div \sqrt{3 } ) + \cos { }^{ - 1} ( - 1 \div 2)$
​

Answer 1

Given Equation

⇒sec⁻¹(2/√3) + cos⁻¹(-1/2)

Using inverse trigonometric function property

⇒cos⁻¹(-x) = π - cos⁻¹(x)

We get

⇒sec⁻¹(2/√3) +π - cos⁻¹(1/2)

we know that

⇒sec(30°) = 2/√3

⇒sec⁻¹(2/√3) = 30°

Similarly

⇒cos(60°) = 1/2

⇒cos⁻¹(1/2) = 60°

Now put the value

⇒30° + π - 60°

We can write π = 180°

⇒30° + 180° - 60°

⇒30° + 120°

⇒150°

Answer

⇒150°

Answer 2

Step-by-step explanation:

ANSWER ✍️

Given Equation

⇒sec⁻¹(2/√3) + cos⁻¹(-1/2)

Using inverse trigonometric function property

⇒cos⁻¹(-x) = π - cos⁻¹(x)

We get

⇒sec⁻¹(2/√3) +π - cos⁻¹(1/2)

we know that

⇒sec(30°) = 2/√3

⇒sec⁻¹(2/√3) = 30°

Similarly

⇒cos(60°) = 1/2

⇒cos⁻¹(1/2) = 60°

Now put the value

⇒30° + π - 60°

We can write π = 180°

⇒30° + 180° - 60°

⇒30° + 120°

⇒150°

Answer

⇒150°

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