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 variable $x$ satisfying the equation $|\sin x \cos x| + \sqrt{2 + \tan^2 x + \cot^2 x} = \sqrt{3}$ belongs to the interval

$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = $

If $a \sin^2 \theta + b \cos^2 \theta = c$,then $\tan^2 \theta$ is equal to

If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$,then the value of $\frac{27 \operatorname{cosec}^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:

If $0 \le x \le \pi$ and $81^{\sin^2 x} + 81^{\cos^2 x} = 30$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module