Prove that:
$\frac{\tan \theta}{\sec \theta-1}=\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}$

Given :

The given statement is $\frac{\tan \theta}{\sec \theta-1}=\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}$.

To do :

We have to prove that $LHS=RHS$ in the given statement.

Solution :

$LHS =\frac{\tan \theta}{\sec \theta-1}$

We know that,

$1+tan^2 \theta = sec^2 \theta$

$tan^2 \theta = sec^2 \theta-1$

$tan^2 \theta = (sec \theta-1)(sec \theta+1)$               $[a^2-b^2=(a+b)(a-b)]$

$tan \theta . tan \theta = (sec \theta-1)(sec \theta+1)$

$\frac{tan \theta}{sec \theta-1}=\frac{sec \theta+1}{tan \theta }$

By ratio theorem,

$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$

So, $\frac{tan \theta}{sec \theta-1}=\frac{sec \theta+1}{tan \theta }= \frac{tan \theta+sec \theta+1}{sec \theta-1+tan \theta}$

Therefore, $\frac{tan \theta}{sec \theta-1}= \frac{tan \theta+sec \theta+1}{tan \theta+sec \theta-1}= RHS$.

Hence proved.

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Updated on: 10-Oct-2022

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