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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 \

Vedclass · Accepted Answer

a natural which is not composite

Vedclass · Answer

(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

Similar Questions

The value of $\sin ^6(\theta) + \cos ^6(\theta) + 3 \sin ^2(\theta) \cos ^2(\theta)$ is

If $\tan^2 \alpha \tan^2 \beta + \tan^2 \beta \tan^2 \gamma + \tan^2 \gamma \tan^2 \alpha + 2\tan^2 \alpha \tan^2 \beta \tan^2 \gamma = 1$,then the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ is

If $\frac{5 \sinh 2x}{7+6 \cosh 2x} = \frac{3}{2}$,then $3 \tanh^2 x + 20 \tanh x = $

$\cos 12^{\circ} \cdot \cos 24^{\circ} \cdot \cos 36^{\circ} \cdot \cos 48^{\circ} \cdot \cos 72^{\circ} \cdot \cos 84^{\circ} = $

$\frac{\cos 12^{\circ}-\sin 12^{\circ}}{\cos 12^{\circ}+\sin 12^{\circ}}+\frac{\sin 147^{\circ}}{\cos 147^{\circ}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module