Answer
To prove
Work Step by Step
It is given that:- \[x=\ln (\sec\theta+\tan\theta)\;\;\;...(1)\]
To show that:- \[\sec \theta=\cosh x\]
Using (1)
\[x=\ln (\sec\theta+\tan\theta)\Rightarrow e^x=\sec\theta+\tan\theta\;\;\;...(2)\]
\[\frac{1}{e^x}=e^{-x}\]
Using (2)
\[\Rightarrow e^{-x}=\frac{1}{\sec\theta+\tan\theta}\]
\[\Rightarrow e^{-x}=\frac{\sec\theta-\tan\theta }{\sec^2\theta-\tan^2\theta}\]
\[\Rightarrow e^{-x}=\frac{\sec\theta-\tan\theta }{1}\]
\[\Rightarrow e^{-x}=\sec\theta-\tan\theta \;\;\;...(3)\]
\[\cosh x=\frac{e^x+e^{-x}}{2}\]
From (2) and (3)
\[\cosh x=\frac{(\sec\theta+\tan\theta)+(\sec\theta-\tan\theta)}{2}=\sec\theta\]
Hence \[\cosh x=\sec\theta\]
