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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 118 of 150

Marks: +1, -0

\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=

Solution:

Concept:

Convert each inverse trigonometric ratio into its tangent value, then apply the tangent addition formula.

Explanation:

Let

A=\sin^{-1}\left(\frac{12}{13}\right),\quad B=\cos^{-1}\left(\frac{4}{5}\right),\quad C=\tan^{-1}\left(\frac{63}{16}\right).

Using right triangles,

\sin A=\frac{12}{13}\Rightarrow \tan A=\frac{12}{5},

\cos B=\frac{4}{5}\Rightarrow \tan B=\frac{3}{4}.

Now apply

\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.

Substitute the values:

\tan(A+B)=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}=-\frac{63}{16}.

Since

A

and

B

are acute and

\tan(A+B)<0

A+B

lies in the second quadrant.

Also,

C

is acute and

\tan C=\frac{63}{16}

. Therefore,

\tan(A+B)=-\tan C.

Thus,

A+B=\pi-C.

So,

A+B+C=\pi.

Answer:

\pi

Option C.

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