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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 58

Answers (1)

Answer:  \frac{2}{5}

Given:

\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0

Hint: Try to convert the separate \sin , \cos  function. So, that the variable get free.

Solution:          

\begin{array}{l} \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0 \\\\ \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=\cos \left(\frac{\pi}{2}\right) \\\\ \sin ^{-1} \frac{2}{5}+\cos ^{-1} x=\frac{\pi}{2} \\\\ x=\frac{2}{5} c g f y \end{array}

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