Question

If $\sin t+\cos t=\frac{1}{5},$ then $\tan \frac{t}{2}$ is equal to:

• A. $-1$
• B. $-\frac{1}{3}$
• C. $2$
• D. $-\frac{1}{6}$

Answer 1

Answer: (B), (C)

The correct options are
B $-\frac{1}{3}$
C $2$

We have,
$\sin t=\frac{2\tan \left(\frac{t}{2}\right)}{1+{\tan }^2\frac{t}{2}}$
$\cos t=\frac{1-{\tan }^2\left(\frac{t}{2}\right)}{1+{\tan }^2\frac{t}{2}}$
Let $y=\tan (\frac{t}{2})$

$\sin t+\cos t=\frac{1}{5}$ becomes

$\frac{2y}{1+y^2}+\frac{1+y^2}{1-y^2}=\frac{1}{5}$

$2y+1-y^2=5+5y^2$
$\Rightarrow (3y+1)(y-2)=0\Rightarrow y=2$ or $\frac{-1}{3}$
Options B, C are correct