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  • need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (iii)

need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (iii)

Answers (1)

Answer:  \frac{1}{2} \cot ^{-1} x

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

Given:   \tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}, x \in R

 

Explanation:

Let            x=\cot \theta ; \theta=\cot ^{-1} x

Now,

                \tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}=\tan ^{-1}\left\{\sqrt{1+\cot ^{2} \theta}-\cot \theta\right\}

As we know, 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta

                \begin{aligned} &\tan ^{-1}\left\{\sqrt{\operatorname{cosec}^{2} \theta}-\cot \theta\right\} \\ &\tan ^{-1}\{\operatorname{cosec} \theta-\cot \theta\} \end{aligned}

As we know, \cot \theta=\frac{\cos \theta}{\sin \theta}, \operatorname{cosec} \theta=\frac{1}{\sin \theta}

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

                \\ \\ \hspace{0.5cm}\because 1-\cos x=2\sin^2\frac{x}{2}\\ \\ \hspace{0.5cm}\because \sin x=2sin \frac{x}{2} \cdot \cos \frac{x}{2}

                \begin{aligned} &\tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta }{2}}\right\} \\ &\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, the principal range of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

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

                \begin{array}{ll} \Rightarrow & \frac{\theta}{2} \end{array}

                \Rightarrow \; \; \; \; \frac{1}{2} \cot ^{-1} x

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