In a triangle ABC,if cos A cos B + sin A sin | vedclass

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Question

(A) Given,$\cos A \cos B + \sin A \sin B \sin C = 1$.We can rewrite this as $\cos A \cos B + \sin A \sin B - \sin A \sin B + \sin A \sin B \sin C = 1$

Vedclass · Accepted Answer

$1 : 1 : \sqrt{2}$

Vedclass · Answer

(A) Given,$\cos A \cos B + \sin A \sin B \sin C = 1$.We can rewrite this as $\cos A \cos B + \sin A \sin B - \sin A \sin B + \sin A \sin B \sin C = 1$.This simplifies to $\cos(A - B) - \sin A \sin B(1 - \sin C) = 1$.Rearranging gives $1 - \cos(A - B) + \sin A \sin B(1 - \sin C) = 0$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we get $2 \sin^2(\frac{A - B}{2}) + \sin A \sin B(1 - \sin C) = 0$.Since the sum of two non-negative terms is zero,each term must be zero: $\sin(\frac{A - B}{2}) = 0$ and $\sin A \sin B(1 - \sin C) = 0$.This implies $A = B$ and $\sin C = 1$ (as $\sin A, \sin B 
eq 0$ in a triangle).Thus,$C = 90^{\circ}$ and $A = B = 45^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 90^{\circ}}$.$a : b : c = \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1 = 1 : 1 : \sqrt{2}$.

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $a : b : c =$

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