Question

If $(sec\theta -tan\theta )=\frac{1}{3}$, the value of $(sec\theta +tan\theta )$ is _____________.

• A. 1
• B. $\sqrt{2}$
• C. 3
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C 3

We know that , $se{c}^2\theta -ta{n}^2\theta =1$ and
$a^2-b^2$ = (a + b) (a - b).

Therefore,
$se{c}^2\theta -ta{n}^2\theta =(sec\theta +tan\theta )(sec\theta -tan\theta )$

So, $(sec\theta +tan\theta )(sec\theta -tan\theta )=1$

Substituiting, $(sec\theta -tan\theta )=\frac{1}{3}$ in the above equation we get,

$\frac{1}{3}\times (sec\theta +tan\theta )=1(sec\theta +tan\theta )=3$