If tan 20^{circ} = k, then frac{tan 250^{circ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $	an 20^{\circ} = k.$We need to evaluate the expression: $\frac{	an 250^{\circ} + 	an 340^{\circ}}{	an 200^{\circ} - 	an 110^{\circ}}.$

Vedclass · Accepted Answer

$\frac{1-k^{2}}{1+k^{2}}$

Vedclass · Answer

(D) Given $	an 20^{\circ} = k.$We need to evaluate the expression: $\frac{	an 250^{\circ} + 	an 340^{\circ}}{	an 200^{\circ} - 	an 110^{\circ}}.$Expressing each term in terms of $20^{\circ}$:$	an 250^{\circ} = 	an(270^{\circ} - 20^{\circ}) = \cot 20^{\circ} = \frac{1}{k}.$$	an 340^{\circ} = 	an(360^{\circ} - 20^{\circ}) = -	an 20^{\circ} = -k.$$	an 200^{\circ} = 	an(180^{\circ} + 20^{\circ}) = 	an 20^{\circ} = k.$$	an 110^{\circ} = 	an(90^{\circ} + 20^{\circ}) = -\cot 20^{\circ} = -\frac{1}{k}.$Substituting these values into the expression:$= \frac{\frac{1}{k} - k}{k - (-\frac{1}{k})} = \frac{\frac{1-k^{2}}{k}}{\frac{k^{2}+1}{k}} = \frac{1-k^{2}}{1+k^{2}}.$

If $\tan 20^{\circ} = k,$ then $\frac{\tan 250^{\circ} + \tan 340^{\circ}}{\tan 200^{\circ} - \tan 110^{\circ}} = $

Similar Questions

If $f(x) = \cos^2 x + \sec^2 x$,then its value is always:

If $\sin x + \sin y = 3(\cos y - \cos x),$ then the value of $\frac{\sin 3x}{\sin 3y}$ is

If $|\cos \theta \{\sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha}\}| \le k$,then the value of $k$ is

The value of $\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 89^{\circ}$ is

In any triangle $ABC$,the expression ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module