Let 0

Question

(B) Given expression: $\sec 2x - 	an 2x = \frac{1 - \sin 2x}{\cos 2x}$ Using identities $\sin 2x = 2 \sin x \cos x$,$\cos 2x = \cos^2 x - \sin^2 x$,a

Vedclass · Accepted Answer

$	an \left( \frac{\pi}{4} - x ight)$

Vedclass · Answer

(B) Given expression: $\sec 2x - 	an 2x = \frac{1 - \sin 2x}{\cos 2x}$ Using identities $\sin 2x = 2 \sin x \cos x$,$\cos 2x = \cos^2 x - \sin^2 x$,and $1 = \cos^2 x + \sin^2 x$: $= \frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{(\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$ $= \frac{\cos x - \sin x}{\cos x + \sin x}$ Dividing numerator and denominator by $\cos x$: $= \frac{1 - 	an x}{1 + 	an x} = \frac{	an \frac{\pi}{4} - 	an x}{1 + 	an \frac{\pi}{4} 	an x}$ Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$: $= 	an \left( \frac{\pi}{4} - x ight)$.

Let $0 < x < \frac{\pi}{4}.$ Then $\sec 2x - \tan 2x = $

Similar Questions

$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot 8\alpha = $

Evaluate: $\sin(22 \frac{1}{2}^{\circ})$

If $\sin^3 x \sin 3x = \sum_{m=0}^n c_m \cos mx$ where $c_0, c_1, c_2, \dots, c_n$ are constants and $c_n \neq 0$,then the value of $n$ is

Evaluate: $\cos^2 A(3 - 4\cos^2 A)^2 + \sin^2 A(3 - 4\sin^2 A)^2$

$\tan 3A \cdot \tan 2A \cdot \tan A = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module