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Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise MCQs Question 9 Maths Textbook Solution.

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Answer: 4 \alpha=3 \beta

Hint: the given question format is like \tan ^{-1}(\tan x).Using this we can solve.

Given: \alpha=\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right), \beta=\tan ^{1}\left(-\tan \frac{2 \pi}{3}\right)


           \begin{aligned} \tan ^{-1}(\tan x) &=x \\ \alpha &=\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right) \\ \alpha &=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{4}\right)\right) \\ \alpha &=\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \\ \alpha &=\frac{\pi}{4} \end{aligned}

                \begin{aligned} &\beta=\tan ^{1}\left(-\tan \frac{2 \pi}{3}\right) \\ &\quad=\tan ^{1}\left(-\tan \left(\pi-\frac{\pi}{3}\right)\right) \\ &\quad=\tan ^{1}\left(\tan \frac{\pi}{3}\right) \end{aligned}

                 \begin{gathered} \beta=\frac{\pi}{3} \\ 4 \alpha=\pi \\ 3 \beta=\pi \end{gathered}

\therefore 4 \alpha=3 \beta

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