Question

If $\sec \theta -\cos \theta =1$, then ${\tan }^2\theta =$

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\frac{\sqrt{7}+1}{2}$
• C. $\frac{\sqrt{5}+1}{3}$
• D. $\frac{\sqrt{5}+1}{2}$

Answer 1

The correct option is D $\frac{\sqrt{5}+1}{2}$

Take $\cos \theta =x$.

So, $\sec \theta =\frac{1}{x}$
$\sec \theta -\cos \theta =\frac{1}{x}-x=1$

$\Rightarrow x^2+x-1=0$

This gives $x=\frac{\sqrt{5}-1}{2}$
so, $\cos \theta =\frac{\sqrt{5}-1}{2}$ and $\sec \theta =\frac{2}{\sqrt{5}-1}$
Rationalising the denominator of $\sec \theta$, we get

$\sec \theta =\frac{\sqrt{5}+1}{2}$

Now, ${\sec }^2\theta =1+{\tan }^2\theta$

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

$\Rightarrow {\tan }^2\theta =\frac{\sqrt{5}+1}{2}$