જો cos A = cos B,,cos C અને A + B + C = pi , | vedclass

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

Question

(d) We have $\cos A = \cos B\cos C$ By triangle property,

$A + B + C = \pi \Rightarrow B + C = \pi - A$

$	herefore \cos (B + C) = \cos (\pi - A

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(d) We have $\cos A = \cos B\cos C$ By triangle property,

$A + B + C = \pi \Rightarrow B + C = \pi - A$

$	herefore \cos (B + C) = \cos (\pi - A) \Rightarrow \cos (B + C) = - \cos A$

$ \Rightarrow \cos B\cos C - \sin B\sin C = - \cos B\cos C$

$( \because {m{Given}}\cos A = \cos B\cos C)$

$ \Rightarrow 2\cos B\cos C = \sin B\sin C$

$ \Rightarrow \frac{{\cos B\cos C}}{{\sin B\sin C}} = \frac{1}{2}$

$\Rightarrow \cot B\cot C = \frac{1}{2}$.

જો $\cos A = \cos B\,\,\cos C$ અને $A + B + C = \pi ,$ તો $\cot \,B\,\cot \,C = . . . ..$

Similar Questions

સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =

જો $A, B, C$ ત્રિકોણના ખૂણા હોય તો $\sin 2A + \sin 2B - \sin 2C$ મેળવો.

$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $

નીચેનામાંથી ક્યાં સમીકરણની કિમત એક થાય

$(sinx + cosecx)^2 + (cosx + secx)^2 - ( tanx + cotx)^2$ =