JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $\sin^{-1} \frac{\alpha}{17} + \cos^{-1} \frac{4}{5} - \tan^{-1} \frac{77}{36} = 0, 0 < \alpha < 13$ , then $\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)$ is equal to:

Choose the correct answer:

Explanation

Solution:

Sabse pehle $\alpha$ nikaalte hain. Maana $\cos^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$ .

\sin^{-1} \frac{\alpha}{17} = \tan^{-1} \frac{77}{36} - \tan^{-1} \frac{3}{4}

\sin^{-1} \frac{\alpha}{17} = \tan^{-1} \left( \frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}} \right) = \tan^{-1} \left( \frac{\frac{77-27}{36}}{\frac{144+231}{144}} \right) = \tan^{-1} \frac{8}{15}

Yadi $\tan \theta = 8/15$ , toh $\sin \theta = 8/17$ . Isliye $\alpha = 8$ .

Ab humein nikalna hai: $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8)$ .

Kyunki $8$ radians lagbhag $2.5\pi$ ke barabar hai ( $8 \approx 2.5 \times 3.14$ ):

$\sin^{-1}(\sin 8) = 3\pi - 8$ (Kyunki $2.5\pi < 8 < 3\pi$ range mein $\sin^{-1}\sin x = (2n+1)\pi - x$ hota hai)
$\cos^{-1}(\cos 8) = 8 - 2\pi$ (Kyunki $2\pi < 8 < 3\pi$ range mein $\cos^{-1}\cos x = x - 2\pi$ hota hai)

Sum: $(3\pi - 8) + (8 - 2\pi) = \pi$

Correct Option: (3) $\pi$

Choose the correct answer:

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