If cot 20^{circ} = p,then frac{tan 160^{circ} | vedclass

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Question

(A) Given expression: $\frac{	an 160^{\circ} - 	an 110^{\circ}}{1 + 	an 160^{\circ} 	an 110^{\circ}}$Using the formula $	an(A - B) = \frac{	an A

Vedclass · Accepted Answer

$\frac{p^{2} - 1}{2p}$

Vedclass · Answer

(A) Given expression: $\frac{	an 160^{\circ} - 	an 110^{\circ}}{1 + 	an 160^{\circ} 	an 110^{\circ}}$Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,the expression simplifies to $	an(160^{\circ} - 110^{\circ}) = 	an 50^{\circ}$.Alternatively,using trigonometric identities:$	an 160^{\circ} = 	an(180^{\circ} - 20^{\circ}) = -	an 20^{\circ} = -\frac{1}{p}$$	an 110^{\circ} = 	an(90^{\circ} + 20^{\circ}) = -\cot 20^{\circ} = -p$Substituting these values into the expression:$= \frac{-	an 20^{\circ} - (- \cot 20^{\circ})}{1 + (-	an 20^{\circ})(-\cot 20^{\circ})}$$= \frac{-\frac{1}{p} + p}{1 + 1} = \frac{\frac{p^{2} - 1}{p}}{2} = \frac{p^{2} - 1}{2p}$

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

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