How to Find the value of a trigonometric function if other very complicated trigonometric equation is given? 
Q)If $\operatorname{sin}\alpha+\operatorname{cos}\alpha=\frac{\sqrt7}2$, $0 \lt \alpha \lt\frac{\pi}{6}, then \operatorname{tan}\frac{\alpha}2 $ is:(1)$\sqrt{7}-2$(2)$(\sqrt{7}-2)/3$(3)$-\sqrt{7}+2$(4)$(-\sqrt{7}+2)/3$

i did this question by two ways but both the was were getting so complex however on solving them completely which took $4.25$ pages the answers were ridiculously different.
$I^{st}$ Method:
squaring both sides $$(\operatorname{sin}\alpha+\operatorname{cos}\alpha)^2=\frac{7}4=1+ \operatorname{sin}2\alpha=\frac 74\implies\operatorname{sin}2\alpha=\frac 34\implies\operatorname{cos}2\alpha=\frac{\sqrt{7}}{4}$$
$$\text{hence, }\operatorname{tan}2\alpha=\frac{3}{\sqrt{7}}\implies 3tan^2\alpha+2\sqrt{7}tan \alpha -3=0\implies tan\alpha=\frac{\surd7 \pm 4}3 $$
no from this if i want to calculate the value of $\tan\alpha/2  $   then you can think how different the answer could be!
$II^{nd} $ Method
$$1+2\sin\alpha\cos\alpha=\frac74\implies \sin^2\alpha\cos^2\alpha=\frac{9}{64}$$
$$64\sin^4\alpha-64\sin^2\alpha+9=0\implies \sin^2\alpha=\frac{64\pm 64\sqrt{7}}{2\times 64}$$
even from this you can have the idea how different the answer will be so different because the value of $$\tan\alpha=\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}$$.

Good news: they are the same!
$$\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}$$
$$=\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}\sqrt{\frac{64 \pm16\surd7}{64 \pm 16\surd7}}$$
$$=\frac{64 \pm16\surd7}{16\sqrt{9}}$$
$$=\frac{4 \pm\surd7}{3}$$
