Question

If $\tan {20}^o=\lambda$, then $\frac{\tan {160}^o-\tan {110}^o}{1+\left(\tan {160}^o\right)\left(\tan {110}^o\right)}=$

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

Answer 1

The correct option is D $\frac{1-{\lambda }^2}{2\lambda }$

If $\tan {20}^{\circ }=\lambda$ , then $\frac{tan{160}^{\circ }-tan{110}^{\circ }}{1+tan{160}^{\circ }tan{110}^{\circ }}=$

As we know that

$tan{160}^{\circ }=tan({180}^{\circ }-{20}^{\circ })$

$tan({180}^{\circ }-{20}^{\circ })=-tan{20}^{\circ }$

Also ,

$tan{110}^{\circ }=tan({90}^{\circ }+{20}^{\circ })$

$tan({90}^{\circ }+{20}^{\circ })=-cot{20}^{\circ }$

Therefore ,

$\frac{tan{160}^{\circ }-tan{110}^{\circ }}{1+tan{160}^{\circ }tan{110}^{\circ }}$

$=\frac{-tan{20}^{\circ }+cot{20}^{\circ }}{1+tan{20}^{\circ }cot{20}^{\circ }}$

Since , $tan{20}^{\circ }=\lambda$

and $cot{20}^{\circ }=\frac{1}{\lambda }$

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

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

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

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