a, b, c are the sides of a scalene triangle A | vedclass

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

Question

(C) Given $\cos \alpha = \frac{a}{b+c}$.Using the identity $\cos \alpha = \frac{1 - 	an^2(\alpha/2)}{1 + 	an^2(\alpha/2)}$,we have $\frac{1 - 	an^2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given $\cos \alpha = \frac{a}{b+c}$.Using the identity $\cos \alpha = \frac{1 - 	an^2(\alpha/2)}{1 + 	an^2(\alpha/2)}$,we have $\frac{1 - 	an^2(\alpha/2)}{1 + 	an^2(\alpha/2)} = \frac{a}{b+c}$.Applying componendo and dividendo,$\frac{(1 + 	an^2(\alpha/2)) + (1 - 	an^2(\alpha/2))}{(1 + 	an^2(\alpha/2)) - (1 - 	an^2(\alpha/2))} = \frac{(b+c) + a}{(b+c) - a}$.This simplifies to $\frac{2}{2 	an^2(\alpha/2)} = \frac{a+b+c}{b+c-a}$,so $	an^2(\alpha/2) = \frac{b+c-a}{a+b+c}$.Similarly,$	an^2(\beta/2) = \frac{c+a-b}{a+b+c}$ and $	an^2(\gamma/2) = \frac{a+b-c}{a+b+c}$.Summing these,$	an^2(\alpha/2) + 	an^2(\beta/2) + 	an^2(\gamma/2) = \frac{(b+c-a) + (c+a-b) + (a+b-c)}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.

$a, b, c$ are the sides of a scalene triangle $ABC$. If angles $\alpha, \beta, \gamma$ lie between $0$ and $\pi$ such that $\cos \alpha = \frac{a}{b+c}, \cos \beta = \frac{b}{c+a}$ and $\cos \gamma = \frac{c}{a+b}$,then $\tan^2 \frac{\alpha}{2} + \tan^2 \frac{\beta}{2} + \tan^2 \frac{\gamma}{2} =$

Similar Questions

If the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$ has real solutions for $\theta$,then $\lambda$ lies in the interval

If $A + B + C = 270^o,$ then $\cos 2A + \cos 2B + \cos 2C + 4\sin A \sin B \sin C = $

If $A + B + C = 180^o$,then $\sum \tan \frac{A}{2} \tan \frac{B}{2} = $

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

The expression $sin^{2n}x + cos^{2n}x$ lies between which of the following values?

Vedclass Test Series

Exam Paper Generator

Online Exam Module