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  • Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 28 math

Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 28 math

Answers (1)

Answer: \pi

Given: 2 \sin ^{-1} \frac{1}{2}+\cos ^{-1}\left(-\frac{1}{2}\right)

Hint: Try to solve the \sin ^{-1} \frac{1}{2} \ \ \& \ \ \cos ^{-1} \frac{1}{2}  and add them.

Solution:

\begin{aligned} 2 \sin ^{-1} \frac{1}{2}+\cos ^{-1}\left(-\frac{1}{2}\right) &=\sin ^{-1} 2 \times \frac{1}{2} \sqrt{1-\left(\frac{1}{2}\right)^{2}}+\cos ^{-1}\left(-\frac{1}{2}\right) \\ &=\sin ^{-1} \frac{\sqrt{3}}{2}+\cos ^{-1}\left(-\frac{1}{2}\right) \end{aligned}

                                                            \begin{array}{l} =\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) \\\\ =\frac{\pi}{3}+\frac{2 \pi}{3} \\\\ =\pi \end{array}

 

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