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Question Number 182842 by malwan last updated on 15/Dec/22

$$provethat$$$$sec\left({tan}^{-1}x\right)=\sqrt{x^2+1}$$

Answered by cortano1 last updated on 15/Dec/22

$$letx=\tan t$$$$\sqrt{x^2+1}=\sqrt{\tan {}^{2}t+1}=\sqrt{\sec {}^{2}t}=\sec t$$$$\Rightarrow \sec \left({\tan }^{-1}\left(\tan t\right)\right)=\sec t$$

Commented by malwan last updated on 15/Dec/22

$$thankyousir$$

Answered by BaliramKumar last updated on 15/Dec/22

$$LHS=sec\left({tan}^{-1}x\right)$$$$LHS=\sqrt{{sec}^2\left({tan}^{-1}x\right)}$$$$LHS=\sqrt{{tan}^2\left({tan}^{-1}x\right)+1}$$$$LHS=\sqrt{x^2+1}=RHS$$$$$$$$$$$$$$

Commented by malwan last updated on 15/Dec/22

$$thankyousomuch$$

Answered by manxsol last updated on 16/Dec/22

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