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

The correct option is D

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

We know that, $se{c}^2\theta$ - $ta{n}^2\theta$ = 1

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

Since, ($sec\theta$ + $tan\theta$) = x

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

Solving both equations

we get tan θ = $\frac{(x^2-1)}{2x}$