$cos(\pi /6 -cos^{-1}(-\sqrt{3}/2))$ is

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

$cos(\pi /6 -cos^{-1}(-\sqrt{3}/2))$ is equal to :

Options:

$-\sqrt{3}/2$

$-1/\sqrt{3}$

$-1/2$

$+1/2$

Correct Answer:

$-1/2$

Explanation:

The correct answer is Option (3) → $-1/2$

$\cos(\pi /6 -\cos^{-1}(-\sqrt{3}/2))$

$=\cos(\pi /6 -\pi +\cos^{-1}\sqrt{3}/2)$

$=\cos(\pi /6 -\pi +\pi /6)$

$=\cos(\pi /3-\pi)=-1/2$