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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 $$\tan\alpha=\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}$$.

anni saini
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3 Answers3

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If $t=\tan⁡\fracα2$ then $$ \cosα = \frac{1-t^2}{1+t^2}\text{ and }\sinα = \frac{2t}{1+t^2} $$ so that $$ a=\cosα+\sinα=\frac{1-t^2+2t}{1+t^2}\iff (1+a)t^2 - 2t +a-1=0\\ \iff ((a+1)t-1)^2 -1 + (a^2-1) \\ \iff t=\frac{1\pm\sqrt{2-a^2}}{a+1}=\frac{a-1}{1\mp\sqrt{2-a^2}} $$ and with $a=\frac{\sqrt7}2$ $$ t=\frac{2a-2}{2\mp\sqrt{8-(2a)^2}}=\frac{\sqrt7-2}{2\pm 1}. $$

Lutz Lehmann
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You need the tangent half angle formula. It says that $\tan(\alpha/2)=\frac{1-\cos \theta}{\sin \theta}$. The link has some other versions.

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

Paul
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