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This is what I tried

$\sec A=\frac{1}{\cos A}$, so the equation becomes

$1-\cos^2A=\cos A$

If we solve the above quadratic equation, we the values of $\cos A$ as $\frac{-1\pm \sqrt5}{2}$

Therefore, $\tan\frac A2$ becomes

$$\sqrt \frac{3-\sqrt 5}{1+\sqrt 5}$$

Squaring that value, the answer remains meaningless

The options are

A) $\sqrt 5+ 2$

B) $\sqrt 5-2$

C) $2-\sqrt5$

D) $0$

Since the options are not matching, where am I going wrong?

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Just notice that $$ \frac{3-\sqrt 5}{1+\sqrt 5}=\frac{(3-\sqrt 5)(1-\sqrt{5})}{(1+\sqrt 5)(1-\sqrt{5})}=\frac{-4\sqrt{5}+8}{-4}=\sqrt{5}-2. $$

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  • $\begingroup$ Dumb mistake on my part. Thanks! $\endgroup$ – Aditya Aug 10 at 16:41
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    $\begingroup$ You are welcome! This can happen quite easily. $\endgroup$ – Dietrich Burde Aug 10 at 16:51
  • $\begingroup$ I actually thought of that, but my mental calculations were messed up so I rejected that theory immediately $\endgroup$ – Aditya Aug 10 at 16:52
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Let $t=\tan^2\dfrac A2$. Then $\cos A=\dfrac{1-t}{1+t}$.

$\dfrac{1+t}{1-t}-\dfrac{1-t}{1+t}=1$

$4t=1-t^2$

$(t+2)^2=5$

$t=-2+\sqrt5$

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