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  • Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (v) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (v) maths textbook solution

Answers (1)

Answer:  \frac{\pi}{2}-\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:

Put        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]            

            \begin{aligned} &\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) \end{aligned}

            \tan ^{-1}\left[\frac{2 \cos ^{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 \cos ^{2} \frac{\theta}{2} \end{array}\right]

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

            \begin{aligned} &\therefore \tan \theta=\cot \left(\frac{\pi}{2}-\theta\right) \\ &\tan ^{-1}\left(\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \end{aligned}

As we know the range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

       \begin{aligned} & \tan ^{-1}(\tan x)=x, x \in\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\} \\ \Rightarrow & \frac{\pi}{2}-\frac{\theta}{2} \\ \Rightarrow & \frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x \end{aligned}

 

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