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

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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}} = $

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