(sec 2A + 1){sec ^2}A = | vedclass

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

Question

(D) Given expression: $(\sec 2A + 1){\sec ^2}A$We know that $\sec 2A = \frac{1}{\cos 2A} = \frac{1 + 	an^2 A}{1 - 	an^2 A}$ and $\sec^2 A = 1 + 	an

Vedclass · Accepted Answer

$2\sec 2A$

Vedclass · Answer

(D) Given expression: $(\sec 2A + 1){\sec ^2}A$We know that $\sec 2A = \frac{1}{\cos 2A} = \frac{1 + 	an^2 A}{1 - 	an^2 A}$ and $\sec^2 A = 1 + 	an^2 A$.Substituting these into the expression:$= \left( \frac{1 + 	an^2 A}{1 - 	an^2 A} + 1 ight) (1 + 	an^2 A)$$= \left( \frac{1 + 	an^2 A + 1 - 	an^2 A}{1 - 	an^2 A} ight) (1 + 	an^2 A)$$= \left( \frac{2}{1 - 	an^2 A} ight) (1 + 	an^2 A)$$= 2 \cdot \frac{1 + 	an^2 A}{1 - 	an^2 A}$Since $\cos 2A = \frac{1 - 	an^2 A}{1 + 	an^2 A}$,it follows that $\sec 2A = \frac{1 + 	an^2 A}{1 - 	an^2 A}$.Therefore,the expression becomes $2 \sec 2A$.

$(\sec 2A + 1){\sec ^2}A = $

Similar Questions

If $b \sin \alpha = a \sin (\alpha + 2\beta)$,then $\frac{a + b}{a - b} = $

If $\tan A + \sin A = m$ and $\tan A - \sin A = n,$ then $\frac{(m^2 - n^2)^2}{mn} = $

If $p = \frac{2 \sin \theta}{1 + \cos \theta + \sin \theta}$ and $q = \frac{\cos \theta}{1 + \sin \theta}$,then

If $\tan \theta + \cot \theta = 5,$ then $\tan^2 \theta + \cot^2 \theta$ is

The value of $\cos^{3}\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{3\pi}{8}\right) + \sin^{3}\left(\frac{\pi}{8}\right) \cdot \sin\left(\frac{3\pi}{8}\right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module