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) $\frac{x^{2}-1}{2 x}$
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).
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