If x sin 45^circ cos^2 60^circ = frac{tan^2 6 | vedclass

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

Question

(C) Given the equation: $x \sin 45^\circ \cos^2 60^\circ = \frac{	an^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$Substitute the trigonom

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Given the equation: $x \sin 45^\circ \cos^2 60^\circ = \frac{	an^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$Substitute the trigonometric values: $\sin 45^\circ = \frac{1}{\sqrt{2}}$,$\cos 60^\circ = \frac{1}{2}$,$	an 60^\circ = \sqrt{3}$,$\csc 30^\circ = 2$,$\sec 45^\circ = \sqrt{2}$,$\cot 30^\circ = \sqrt{3}$.$x \cdot \frac{1}{\sqrt{2}} \cdot (\frac{1}{2})^2 = \frac{(\sqrt{3})^2 \cdot 2}{\sqrt{2} \cdot (\sqrt{3})^2}$$x \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{4} = \frac{3 \cdot 2}{\sqrt{2} \cdot 3}$$\frac{x}{4\sqrt{2}} = \frac{6}{3\sqrt{2}}$$\frac{x}{4\sqrt{2}} = \frac{2}{\sqrt{2}}$$x = 2 \cdot 4 = 8$.

If $x \sin 45^\circ \cos^2 60^\circ = \frac{\tan^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$,then $x = $

Similar Questions

If $\alpha + \beta - \gamma = \pi,$ then $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = $

If $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d}$,then $\left( \frac{a + c}{b + d} \right)$ is equal to :-

If $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ ,$ then the value of $x$ is

$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} = $

$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module