If P = tan 20^circ, then the value of frac{ta | vedclass

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(D) We use the identity $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Given expression is $\frac{	an 160^\circ - 	an 110^\circ}{1 + 	an

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

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(D) We use the identity $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Given expression is $\frac{	an 160^\circ - 	an 110^\circ}{1 + 	an 160^\circ 	an 110^\circ} = 	an(160^\circ - 110^\circ) = 	an 50^\circ$.We know $	an 50^\circ = 	an(70^\circ - 20^\circ)$ or $	an(30^\circ + 20^\circ)$. However, using $	an(90^\circ - 40^\circ) = \cot 40^\circ$ is also possible.Let's use $	an 50^\circ = \cot 40^\circ = \frac{1}{	an 40^\circ}$.Using $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$, where $	heta = 20^\circ$:$	an 40^\circ = \frac{2 	an 20^\circ}{1 - 	an^2 20^\circ} = \frac{2P}{1 - P^2}$.Therefore, $	an 50^\circ = \frac{1}{\frac{2P}{1 - P^2}} = \frac{1 - P^2}{2P}$.

If $P = \tan 20^\circ$, then the value of $\frac{\tan 160^\circ - \tan 110^\circ}{1 + \tan 160^\circ \tan 110^\circ}$ in terms of $P$, is...

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