# If $\sec A-\cos A=1$, then determine the value of $\tan^2\frac A2$

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?

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

• Dumb mistake on my part. Thanks! – Aditya Aug 10 at 16:41
• You are welcome! This can happen quite easily. – Dietrich Burde Aug 10 at 16:51
• I actually thought of that, but my mental calculations were messed up so I rejected that theory immediately – Aditya Aug 10 at 16:52

Let $$t=\tan^2\dfrac A2$$. Then $$\cos A=\dfrac{1-t}{1+t}$$.

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

$$4t=1-t^2$$

$$(t+2)^2=5$$

$$t=-2+\sqrt5$$