Question

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QUESTION IS IN THE ATTACHMENT.✔️


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Answer 1

$\huge \texttt{Proof...}$

$\Large \textsf{LHS...}$

$\displaystyle \Longrightarrow\ \ \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} \\ \\ \\ \Longrightarrow\ \ \frac{(\sin\theta-\cos\theta+1)\sec\theta}{(\sin\theta+\cos\theta-1)\sec\theta} \\ \\ \\ \Longrightarrow\ \ \frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta} \\ \\ \\ \Longrightarrow\ \ \frac{(\tan\theta+\sec\theta-1)(\sec\theta-\tan\theta)}{(\tan\theta-\sec\theta+1)(\sec\theta-\tan\theta)}$

$\displaystyle \Longrightarrow\ \ \frac{(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)-(\sec\theta-\tan\theta)}{(\tan\theta-\sec\theta+1)(\sec\theta-\tan\theta)} \\ \\ \\ \Longrightarrow\ \ \frac{\sec^2\theta-\tan^2\theta-\sec\theta+\tan\theta}{(\tan\theta-\sec\theta+1)(\sec\theta-\tan\theta)} \\ \\ \\ \Longrightarrow\ \ \frac{\tan\theta-\sec\theta+1}{(\tan\theta-\sec\theta+1)(\sec\theta-\tan\theta)} \\ \\ \\ \Longrightarrow\ \ \frac{1}{\sec\theta-\tan\theta}$

$\Longrightarrow\ \ \Large \textsf{...RHS}$

$\Huge \textsc{\underline{\underline{Hence Proved!!!}}}$