if secθ+tanθ=x, then tanθ=
Question:

If $\sec \theta+\tan \theta=x$, then $\tan \theta=$

(a) $\frac{x^{2}+1}{x}$

(b) $\frac{x^{2}-1}{x}$

(c) $\frac{x^{2}+1}{2 x}$

(d) $\frac{x^{2}-1}{2 x}$

Solution:

Given:

$\sec \theta+\tan \theta=x$

We know that,

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

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

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

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

Now,

$\sec \theta+\tan \theta=x$

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

Subtracting the second equation from the first equation, we get

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

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

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

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

Therefore, the correct choice is (d).