The expression frac{tan(x - frac{pi}{2}) cdot | vedclass

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

Question

(D) Using trigonometric identities:$1$. $	an(x - \frac{\pi}{2}) = -\cot x$$2$. $\cos(\frac{3\pi}{2} + x) = \sin x$$3$. $\sin(\frac{7\pi}{2} - x) = \s

Vedclass · Accepted Answer

$\sin^2 x$

Vedclass · Answer

(D) Using trigonometric identities:$1$. $	an(x - \frac{\pi}{2}) = -\cot x$$2$. $\cos(\frac{3\pi}{2} + x) = \sin x$$3$. $\sin(\frac{7\pi}{2} - x) = \sin(4\pi - \frac{\pi}{2} - x) = \sin(-\frac{\pi}{2} - x) = -\cos x$$4$. $\cos(x - \frac{\pi}{2}) = \sin x$$5$. $	an(\frac{3\pi}{2} + x) = -\cot x$Substituting these into the expression:Numerator: $(-\cot x)(\sin x) - (-\cos x)^3 = -\cos x + \cos^3 x = \cos x(\cos^2 x - 1) = -\cos x \sin^2 x$Denominator: $(\sin x)(-\cot x) = -\cos x$Expression: $\frac{-\cos x \sin^2 x}{-\cos x} = \sin^2 x$.

The expression $\frac{\tan(x - \frac{\pi}{2}) \cdot \cos(\frac{3\pi}{2} + x) - \sin^3(\frac{7\pi}{2} - x)}{\cos(x - \frac{\pi}{2}) \cdot \tan(\frac{3\pi}{2} + x)}$ simplifies to:

Similar Questions

The value of $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ} = $

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

If $\tan 3^{\circ} + 2\tan 6^{\circ} + 4\tan 12^{\circ} + 8\cot 24^{\circ} = \cot \theta^{\circ}$,then:

$ABC$ is a triangle. If $\sin \left(\frac{A+B}{2}\right) = \frac{\sqrt{3}}{2}$,then the value of $\sin \frac{C}{2}$ is

If $\Delta ABC$ is right-angled at $B$,$AB = 6$ units,and $\angle C = 30^{\circ}$,then $AC$ is equal to (in units):

Vedclass Test Series

Exam Paper Generator

Online Exam Module