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

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

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Answer: \frac{1}{2} \tan ^{-1} x

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

Given: \tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\}, x \neq 0

Explanation:

Let       x=\tan \theta, \text { then } \theta=\tan ^{-1} x

          \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)  

As we know, 1+\tan ^{2} \theta=\sec ^{2} \theta

            \begin{aligned} &\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right) \\ &\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \end{aligned}

            \tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) \quad \because\left[\sec \theta=\frac{1}{\cos \theta}, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]

            \tan ^{-1}\left(\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right)

            \tan ^{-1}\left(\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right)

            \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)

            \tan ^{-1}\left[\frac{2 \sin^{2} \frac{ \theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \quad \because\left[\begin{array}{c} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2} \end{array}\right]

            \begin{aligned} &\tan ^{-1}\left[\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right] \\ &\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \end{aligned}

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

            \tan ^{-1}\left[\tan \frac{\theta}{2}\right]=\frac{\theta}{2}

Now putting the value of θ

    \Rightarrow \quad \frac{1}{2} \tan ^{-1} x

 

Posted by

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