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

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

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a natural which is not composite

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(D) Given expression: $E = \frac{	an^2 20^\circ - \sin^2 20^\circ}{	an^2 20^\circ \cdot \sin^2 20^\circ}$.We know that $	an^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta}$.Numerator: $	an^2 20^\circ - \sin^2 20^\circ = \frac{\sin^2 20^\circ}{\cos^2 20^\circ} - \sin^2 20^\circ = \sin^2 20^\circ \left( \frac{1}{\cos^2 20^\circ} - 1 ight)$.Using $1 - \cos^2 	heta = \sin^2 	heta$,we get $\frac{1 - \cos^2 20^\circ}{\cos^2 20^\circ} = \frac{\sin^2 20^\circ}{\cos^2 20^\circ} = 	an^2 20^\circ$.So,the numerator becomes $\sin^2 20^\circ \cdot 	an^2 20^\circ$.Substituting this back into the expression: $E = \frac{\sin^2 20^\circ \cdot 	an^2 20^\circ}{	an^2 20^\circ \cdot \sin^2 20^\circ} = 1$.Since $1$ is a natural number and it is neither prime nor composite,the correct description is that it is a natural number which is not composite.

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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