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

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

Answers (1)

Answer:   \frac{\pi}{2}-\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\{x+\sqrt{1+x^{2}}\right\}, x \in R

Explanation:

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

Now ,

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

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

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

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

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

∴  As we know, \sin 2 \theta=2 \sin \theta \cos \theta

            \begin{aligned} &1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \\ &\tan ^{-1}\left\{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right\} \end{aligned}

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

  As we know \tan \theta=\cot \left(\frac{\pi}{2}-\theta\right)

            \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}

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

\Rightarrow \quad\left(\frac{\pi}{2}-\frac{\theta}{2}\right)

Now put value of θ

\Rightarrow \quad \frac{\pi}{2}-\frac{\cot ^{-1} x}{2}

 

Posted by

infoexpert26

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