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  • explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (viii) maths

explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (viii) maths

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Answer:  \frac{\pi}{4}+\sin ^{-1} x

Hint:  The range of principal value of \sin ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given:  \sin ^{-1}\left[\frac{x+\sqrt{1+x^{2}}}{\sqrt{2}}\right], \frac{1}{2}<x<\frac{1}{\sqrt{2}}

Explanation:

Let         x=\sin \alpha

Then      \alpha=\sin ^{-1} x

            \begin{aligned} &\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}=\frac{\sin \alpha+\sqrt{1-\sin ^{2} \alpha}}{\sqrt{2}} \\ &\therefore \sin ^{2} \alpha+\cos ^{2} \alpha=1 \end{aligned}

Then,     1-\sin ^{2} \alpha=\cos ^{2} \alpha

            \begin{aligned} &=\frac{\sin \alpha+\sqrt{\cos ^{2} \alpha}}{\sqrt{2}} \\ &=\frac{\sin \alpha+\sqrt{\cos ^{2} \alpha}}{\sqrt{2}} \end{aligned}

            \begin{aligned} &=\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}+\cos \alpha \cdot \frac{1}{\sqrt{2}}\right) \\ &\therefore \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \end{aligned}

            \begin{aligned} &=\sin \alpha \cdot \cos \left(\frac{\pi}{4}\right)+\cos \alpha \cdot \sin \left(\frac{\pi}{4}\right) \\ &\therefore \sin A \cos B+\cos A \sin B=\sin (A+B) \end{aligned}

            \sin \alpha \cos \left(\frac{\pi}{4}\right)+\cos \alpha \sin \left(\frac{\pi}{4}\right)=\sin \left(\alpha+\frac{\pi}{4}\right)

Then put value of    \frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}=\sin \left(\alpha+\frac{\pi}{4}\right)

            \sin ^{-1}\left(\sin \left(\alpha+\frac{\pi}{4}\right)\right)

As we know,   \sin ^{-1}(\sin x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

            \begin{aligned} &=\alpha+\frac{\pi}{4} \\ &=\sin ^{-1} x+\frac{\pi}{4} \end{aligned}

 

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