The value of sin ^{2} 22^{circ}+sin ^{2} 68^{ | vedclass

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

Question

(B) Given expression: $\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}+\cot ^{2} 30^{\circ}$We know that $\sin(90^{\circ}-	heta) = \cos 	heta$.Therefore,$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given expression: $\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}+\cot ^{2} 30^{\circ}$We know that $\sin(90^{\circ}-	heta) = \cos 	heta$.Therefore,$\sin 22^{\circ} = \sin(90^{\circ}-68^{\circ}) = \cos 68^{\circ}$.Substituting this into the expression,we get:$\cos ^{2} 68^{\circ} + \sin ^{2} 68^{\circ} + \cot ^{2} 30^{\circ}$Using the identity $\sin ^{2} 	heta + \cos ^{2} 	heta = 1$,we have:$1 + \cot ^{2} 30^{\circ}$Since $\cot 30^{\circ} = \sqrt{3}$,then $\cot ^{2} 30^{\circ} = (\sqrt{3})^{2} = 3$.Thus,the value is $1 + 3 = 4$.

The value of $\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}+\cot ^{2} 30^{\circ}$ is:

Similar Questions

If $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5},$ where $\theta$ and $\phi$ are positive acute angles,then $\cos \frac{\theta - \phi}{2} = $

If $\sin A = n \sin B,$ then $\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = $

The value of $\tan^{-1} \left( \frac{\sin 2 - 1}{\cos 2} \right)$ is equal to:

If $\alpha = 22^\circ 30'$,then $(1 + \cos \alpha )(1 + \cos 3\alpha )(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ equals

The minimum value of $\cos 2\theta + \cos \theta$ for real values of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module