Question

$\sin \alpha +\cos \alpha =1/5$, then: $\tan \left(\alpha /2\right)=$_______.

• A. $1$ or $(-1/2)$
• B. $2$ or $(-1/3)$
• C. $3$ or $(-1/4)$
• D. None of these

Answer 1

Answer: (B)

$2$ or $(-1/3)$

Given

$\sin \alpha +\cos \alpha =\frac{1}{5}$

$⟹\frac{2\tan \frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}+\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{1}{5}$

$⟹10\tan \frac{\alpha }{2}+5-5{\tan }^2\frac{\alpha }{2}=1+{\tan }^2\frac{\alpha }{2}$

$⟹6{\tan }^2\frac{\alpha }{2}-10\tan \frac{\alpha }{2}-4=0$

$⟹3{\tan }^2\frac{\alpha }{2}-5\tan \frac{\alpha }{2}-2=0$

$⟹3{\tan }^2\frac{\alpha }{2}-6\tan \frac{\alpha }{2}+\tan \frac{\alpha }{2}-2=0$

$⟹3\tan \frac{\alpha }{2}(\tan \frac{\alpha }{2}-2)+1(\tan \frac{\alpha }{2}-2)$

$⟹(\tan \frac{\alpha }{2}+\frac{1}{3})(\tan \frac{\alpha }{2}-2)=0$

$⟹\tan \frac{\alpha }{2}=2\left(or\right)-\frac{1}{3}$