Question

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

• A. $\frac{x^2+1}{x}$
• B. $\frac{x^2-1}{x}$
• C. $\frac{x^2+1}{2x}$
• D. $\frac{x^2-1}{2x}$

Answer 1

Answer: (D)

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

Given $\text{sec}\theta +\tan \theta =x\cdots \left(1\right)$

As we know that

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

$⟹(\text{sec}\theta -\tan \theta )(\text{sec}\theta +\tan \theta )=1$ $(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

$⟹\text{sec}\theta -\tan \theta =\frac{1}{x}\cdots \left(2\right)$

$\left(1\right)-\left(2\right)⟹2\tan \theta =x-\frac{1}{x}=\frac{x^2-1}{x}⟹\tan \theta =\frac{x^2-1}{2x}$