The expression frac{tan^2 20^circ - sin^2 20^ | vedclass

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(D) Let the expression be $E = \frac{	an^2 20^\circ - \sin^2 20^\circ}{	an^2 20^\circ \cdot \sin^2 20^\circ}$.We know that $	an^2 	heta - \sin^2 \

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a natural which is not composite

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(D) Let the expression be $E = \frac{	an^2 20^\circ - \sin^2 20^\circ}{	an^2 20^\circ \cdot \sin^2 20^\circ}$.We know that $	an^2 	heta - \sin^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta} - \sin^2 	heta = \sin^2 	heta \left( \frac{1}{\cos^2 	heta} - 1 ight) = \sin^2 	heta \left( \frac{1 - \cos^2 	heta}{\cos^2 	heta} ight) = \sin^2 	heta \cdot \frac{\sin^2 	heta}{\cos^2 	heta} = 	an^2 	heta \cdot \sin^2 	heta$.Substituting $	heta = 20^\circ$,we get $	an^2 20^\circ - \sin^2 20^\circ = 	an^2 20^\circ \cdot \sin^2 20^\circ$.Therefore,$E = \frac{	an^2 20^\circ \cdot \sin^2 20^\circ}{	an^2 20^\circ \cdot \sin^2 20^\circ} = 1$.Since $1$ is a natural number and it is not composite (it is neither prime nor composite),the correct option is $D$.

The expression $\frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ}$ simplifies to

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