Question

If $\sec \theta +\tan \theta =x$ then find tan $\theta .$

Answer 1

$\sec \theta +\tan \theta =x\to \left(1\right)$

Now, we know that

$\Rightarrow {\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow ({\sec }^2\theta -{\tan }^2\theta )(\sec \theta +\tan \theta )=1$

$\Rightarrow (\sec \theta -\tan \theta )x=1$

$\Rightarrow \sec \theta -\tan \theta =\frac{1}{x}\to \left(2\right)$

Subtracting, $\left(1\right)-\left(2\right)$, we get

$\Rightarrow 2\tan \theta =x-\frac{1}{x}$

$\Rightarrow \tan \theta =\frac{1}{2}(x-\frac{1}{x})$

$\Rightarrow \tan \theta =\frac{1}{2x}(x^2-1)$

Hence, the answer is $\frac{1}{2x}(x^2-1).$