If tan 20^{circ}=lambda,then the value of fra | vedclass

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(D) Given,$	an 20^{\circ}=\lambda$. Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,the expression becomes: $\frac{	an 160

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given,$	an 20^{\circ}=\lambda$. Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,the expression becomes: $\frac{	an 160^{\circ}-	an 110^{\circ}}{1+	an 160^{\circ} 	an 110^{\circ}} = 	an(160^{\circ}-110^{\circ}) = 	an 50^{\circ}$. However,we can also simplify the terms individually: $	an 160^{\circ} = 	an(180^{\circ}-20^{\circ}) = -	an 20^{\circ} = -\lambda$. $	an 110^{\circ} = 	an(90^{\circ}+20^{\circ}) = -\cot 20^{\circ} = -\frac{1}{\lambda}$. Substituting these into the expression: $\frac{-\lambda - (-1/\lambda)}{1 + (-\lambda)(-1/\lambda)} = \frac{-\lambda + 1/\lambda}{1 + 1} = \frac{\frac{1-\lambda^2}{\lambda}}{2} = \frac{1-\lambda^2}{2\lambda}$.

If $\tan 20^{\circ}=\lambda$,then the value of $\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\left(\tan 160^{\circ}\right)\left(\tan 110^{\circ}\right)}$ is:

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