2 π-≤ft( ^{-1} 4/5+ ^{-1} 5/13+ ^{-1} 16/65) is equal to

Correct Answer is : 3π23 π/223π

Correct Answer is : 3π23 π/223π

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MHTCET 15 May 2024 Shift 1 Paper

Question : 134 of 150

Marks: +1, -0

2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right)

is equal to

Solution:

Concept:

Convert inverse sine terms into inverse tangent terms, then combine them using the tangent addition formula.

Explanation:

For

\sin^{-1}\frac{4}{5}

, we have

\tan^{-1}\frac{4}{3}

For

\sin^{-1}\frac{5}{13}

, we have

\tan^{-1}\frac{5}{12}

So the given expression becomes:

2\pi - \left[\left(\tan^{-1}\frac{4}{3} + \tan^{-1}\frac{5}{12}\right) + \sin^{-1}\frac{16}{65}\right]

Using

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

\tan^{-1}\frac{4}{3} + \tan^{-1}\frac{5}{12} = \tan^{-1}\left(\frac{\frac43+\frac{5}{12}}{1-\frac43\cdot\frac{5}{12}}\right)

= \tan^{-1}\left(\frac{\frac{21}{12}}{\frac{4}{9}}\right) = \tan^{-1}\frac{63}{16}

Now, if

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

, then

\cos\theta = \frac{16}{65}

Hence,

\tan^{-1}\frac{63}{16} = \cos^{-1}\frac{16}{65}

The expression becomes:

2\pi - \left[\cos^{-1}\frac{16}{65} + \sin^{-1}\frac{16}{65}\right]

Since

\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}

for

0 \le x \le 1

2\pi - \frac{\pi}{2} = \frac{3\pi}{2}

Answer:

2\pi - \left(\sin^{-1}\frac{4}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{16}{65}\right) = \frac{3\pi}{2}

Therefore, the correct option is D. $\frac{3\pi}{2}$ .

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