The value of ^{-1}≤ft({ ≤ft(19 π/4)-1}{ ≤ft(π/4)}) is equal to

Correct Answer is : −3π8-3 π/8−83π

Correct Answer is : −3π8-3 π/8−83π

Test Index

MHTCET 15 Apr 2026 Shift 2 Paper

Question : 120 of 150

Marks: +1, -0

The value of

\tan ^{-1}\left(\frac{\cos \left(\frac{19 \pi}{4}\right)-1}{\sin \left(\frac{\pi}{4}\right)}\right)

is equal to

Solution:

Concept:

Use periodicity of cosine and standard inverse tangent values.

Explanation:

Simplify the angle inside cosine first.

\frac{19\pi}{4} = 4\pi + \frac{3\pi}{4}

Since

\cos

has period

2\pi

\cos\left(\frac{19\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}

Also,

\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

Substitute these values into the expression:

\displaystyle \frac{\cos\left(\frac{19\pi}{4}\right)-1}{\sin\left(\frac{\pi}{4}\right)} = \frac{-\frac{1}{\sqrt{2}} - 1}{\frac{1}{\sqrt{2}}}

Simplify the numerator:

\displaystyle -\frac{1}{\sqrt{2}} - 1 = -\frac{1+\sqrt{2}}{\sqrt{2}}

Dividing by

\frac{1}{\sqrt{2}}

gives:

-(1+\sqrt{2})

So we need

\tan^{-1}(-(1+\sqrt{2}))

We know

\tan\left(\frac{3\pi}{8}\right) = 1+\sqrt{2}

Therefore,

\tan\left(-\frac{3\pi}{8}\right) = -(1+\sqrt{2})

Since

-\frac{3\pi}{8}

lies in the principal branch

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

\tan^{-1}(-(1+\sqrt{2})) = -\frac{3\pi}{8}

Answer:

\displaystyle -\frac{3\pi}{8}

, i.e. Option C.

Go to Question:

Prev Question

Next Question