Question

If $\tan {25}^o=x$, then $\frac{\tan {155}^o-\tan {115}^o}{1+\tan {155}^o\tan {115}^o}$ is equal to?

• A. $\frac{1-x^2}{2x}$
• B. $\frac{1+x^2}{2x}$
• C. $\frac{1+x^2}{1-x^2}$
• D. $\frac{1-x^2}{1+x^2}$

Answer 1

The correct option is A $\frac{1-x^2}{2x}$

$\tan 25°=x$

$\frac{\tan 155°-\tan 115°}{1+\tan 155°\tan 115°}$

$\frac{\tan (180°-25°)-\tan (90°+25°)}{1+\tan (180°-25°)\tan (90°+25°)}$

$\tan (180°-\theta )=-\tan \theta$

$\tan (90°+\theta )=-\cot \theta$

$=\frac{-\tan 25°+\cot 25°}{1(-\tan 25°)(-\cot 25°)}$

$=\frac{-x+\frac{1}{x}}{1+x\times \frac{1}{x}}$

$=\frac{-x^2+1}{2x}$

$=\frac{1-x^2}{2x}$