If 7 sin^{2} theta + 3 cos^{2} theta = 4,then | vedclass

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

Question

(D) Given equation: $7 \sin^{2} 	heta + 3 \cos^{2} 	heta = 4$.Using the identity $\cos^{2} 	heta = 1 - \sin^{2} 	heta$,we substitute:$7 \sin^{2} \

Vedclass · Accepted Answer

$1/\sqrt{3}$

Vedclass · Answer

(D) Given equation: $7 \sin^{2} 	heta + 3 \cos^{2} 	heta = 4$.Using the identity $\cos^{2} 	heta = 1 - \sin^{2} 	heta$,we substitute:$7 \sin^{2} 	heta + 3(1 - \sin^{2} 	heta) = 4$$7 \sin^{2} 	heta + 3 - 3 \sin^{2} 	heta = 4$$4 \sin^{2} 	heta = 4 - 3$$4 \sin^{2} 	heta = 1$$\sin^{2} 	heta = 1/4$Since $	heta$ is acute,$\sin 	heta = 1/2$.Thus,$	heta = 30^{\circ}$.Therefore,$	an 	heta = 	an 30^{circ} = 1/\sqrt{3}$.

If $7 \sin^{2} \theta + 3 \cos^{2} \theta = 4$,then the value of $\tan \theta$ is (where $\theta$ is acute).

Similar Questions

$\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }} = $

$\sqrt {2 + \sqrt {2 + 2\cos 4\theta } } = $

If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,then $x$ must be

If $\sin \theta = -\frac{7}{25}$ and $\theta$ is in the third quadrant,then $\frac{7 \cot \theta - 24 \tan \theta}{7 \cot \theta + 24 \tan \theta} = $

The value of $\sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module