If alpha+beta=frac{pi}{2} and beta+gamma=alph | vedclass

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

Question

(C) Given,$\alpha = \beta + \gamma$. Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$. Substituting $\beta$ in the firs

Vedclass · Accepted Answer

$	an \beta+2 	an \gamma$

Vedclass · Answer

(C) Given,$\alpha = \beta + \gamma$. Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$. Substituting $\beta$ in the first equation: $\alpha = (\frac{\pi}{2} - \alpha) + \gamma$,which implies $\gamma = 2\alpha - \frac{\pi}{2}$. Alternatively,using $\gamma = \alpha - \beta$: $	an \gamma = 	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$. Since $\beta = \frac{\pi}{2} - \alpha$,then $	an \beta = \cot \alpha = \frac{1}{	an \alpha}$. Substituting this: $	an \gamma = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha (\frac{1}{	an \alpha})} = \frac{	an \alpha - 	an \beta}{1 + 1} = \frac{	an \alpha - 	an \beta}{2}$. Therefore,$2 	an \gamma = 	an \alpha - 	an \beta$,which gives $	an \alpha = 	an \beta + 2 	an \gamma$.

If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$,then $\tan \alpha$ equals

Similar Questions

If $A + B + C = \frac{3\pi}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $

The minimum value of $(8 \sec^2 \theta + 2 \cos^2 \theta)$ is equal to :-

The number of integral values of $k$,for which the equation $7 \cos x + 5 \sin x = 2k + 1$ has a solution,is

If $\alpha \in \left( 0, \frac{\pi}{2} \right)$,then $\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$ is always greater than or equal to

The maximum value of the function $f(x) = 3 \sin^{12} x + 4 \cos^{16} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module