If tan left( frac{pi}{4} + theta right) + tan | vedclass

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

Question

(D) We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Using these,$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Using these,$	an \left( \frac{\pi}{4} + 	heta ight) = \frac{1 + 	an 	heta}{1 - 	an 	heta}$ and $	an \left( \frac{\pi}{4} - 	heta ight) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$.Adding these two expressions:$\frac{1 + 	an 	heta}{1 - 	an 	heta} + \frac{1 - 	an 	heta}{1 + 	an 	heta} = \frac{(1 + 	an 	heta)^2 + (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$$= \frac{1 + 	an^2 	heta + 2 	an 	heta + 1 + 	an^2 	heta - 2 	an 	heta}{1 - 	an^2 	heta}$$= \frac{2(1 + 	an^2 	heta)}{1 - 	an^2 	heta}$Since $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$,we have $\sec 2	heta = \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta}$.Thus,the expression becomes $2 \sec 2	heta$.Comparing this with $\lambda \sec 2	heta$,we get $\lambda = 2$.

If $\tan \left( \frac{\pi}{4} + \theta \right) + \tan \left( \frac{\pi}{4} - \theta \right) = \lambda \sec 2\theta$,then $\lambda$ =

Similar Questions

If $A + B + C = \pi$ $(A, B, C > 0)$ and the angle $C$ is obtuse,then:

If $\sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ} = k$,then $k =$

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

If $x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ and $x \sin \alpha - y \cos \alpha = 0$,then $x^2 + y^2 = $

If $\theta$ is in the first quadrant and $\tan \theta = \frac{3}{4},$ then $\frac{\tan \left(\frac{\pi}{2}-\theta\right)-\sin (\pi-\theta)}{\sin \left(\frac{3 \pi}{2}+\theta\right)-\cot (2 \pi-\theta)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module