frac{sin 60^{circ} + cos 30^{circ}}{1 + sin 3 | vedclass

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

Question

(D) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 30^{\circ} = \frac{1}{2}$,an

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(D) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 30^{\circ} = \frac{1}{2}$,and $\cos 60^{\circ} = \frac{1}{2}$.Substituting these values into the expression:$\frac{\sin 60^{\circ} + \cos 30^{\circ}}{1 + \sin 30^{\circ} + \cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} + \frac{1}{2}}$$= \frac{\frac{2\sqrt{3}}{2}}{1 + 1}$$= \frac{\sqrt{3}}{2}$

$\frac{\sin 60^{\circ} + \cos 30^{\circ}}{1 + \sin 30^{\circ} + \cos 60^{\circ}} = \dots$

Similar Questions

Given that $\sin \alpha = \frac{1}{2}$ and $\cos \beta = \frac{1}{2}$,then the value of $(\alpha + \beta)$ is (in $^{\circ}$)

If $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$,then $x = \ldots$ (in $^{\circ}$)

State whether the following is 'True' or 'False' and justify your answer:
If $\cos A + \cos^2 A = 1$,then $\sin^2 A + \sin^4 A = 1$.

For acute angle $\theta,$ if $\cos \theta = \sin \theta,$ then $2 \tan^{2} \theta + \sin^{2} \theta + 1 = \ldots$

$\frac{1}{\cos ^{2} \theta}-1 = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module