If tan A = 2tan B + cot B, then 2tan (A - B) | vedclass

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

Question

(C) Given $	an A = 2	an B + \cot B.$We know that $	an (A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}.$Substituting the value of $	an A$:$2	a

Vedclass · Accepted Answer

$\cot B$

Vedclass · Answer

(C) Given $	an A = 2	an B + \cot B.$We know that $	an (A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}.$Substituting the value of $	an A$:$2	an (A - B) = 2 \left( \frac{2	an B + \cot B - 	an B}{1 + (2	an B + \cot B)	an B} ight)$$= 2 \left( \frac{	an B + \cot B}{1 + 2	an^2 B + \cot B 	an B} ight)$Since $\cot B 	an B = 1$:$= 2 \left( \frac{	an B + \cot B}{1 + 2	an^2 B + 1} ight) = 2 \left( \frac{	an B + \cot B}{2 + 2	an^2 B} ight)$$= 2 \left( \frac{	an B + \cot B}{2(1 + 	an^2 B)} ight) = \frac{	an B + \cot B}{\sec^2 B}$$= \frac{	an B}{\sec^2 B} + \frac{\cot B}{\sec^2 B} = \sin B \cos B + \frac{\cos B}{\sin B} \cos^2 B = \sin B \cos B + \frac{\cos^3 B}{\sin B}$Alternatively,simplifying the expression directly:$= \frac{	an B + \frac{1}{	an B}}{1 + 	an^2 B} = \frac{\frac{	an^2 B + 1}{	an B}}{1 + 	an^2 B} = \frac{1}{	an B} = \cot B.$

If $\tan A = 2\tan B + \cot B,$ then $2\tan (A - B) = $

Similar Questions

If $\sin \theta \cosh \alpha = \tan x$ and $\cos \theta \sinh \alpha = \sec x$,then find the value of $\cos 2 \theta \cosh 2 \alpha$.

If $\theta$ is in the interval $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $\cos 2 \theta \cdot \sec ^4 \theta + \sec ^2 \theta = 0$,then $\sin ^2 \theta =$

If $\tan \theta + \sin \theta = a$ and $\tan \theta - \sin \theta = b$,then the values of $\cot \theta$ and $\operatorname{cosec} \theta$ are respectively

If $\sin (\theta + \alpha ) = a$ and $\sin (\theta + \beta ) = b,$ then $\cos 2(\alpha - \beta ) - 4ab\cos (\alpha - \beta )$ is equal to

If $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$,then the value of the expression $\frac{\sin(A + B)}{\sin 2C}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module