Question

we have ( cos A cos B+sin A sin B sin C=1 )
( Rightarrow cos A cos B+sin A sin B-sin A sin B )
( +sin A sin B sin C=1 )
( Rightarrow cos (A-B)-sin A sin B(1-sin C)=0 )
( Rightarrow 2 sin ^{2}left(frac{A-B}{2}right)+sin A sin B(1-sin C)=0 )
In this term This teem is squase ( A, B, C ) all are term : it is less than ( 180^{circ} ) but positine
( therefore ) Sin ( A ) sinB and ( sin C ) are the and less than 1 . So second term is also positine But RHS = 0 . Which in possible oney if each tesm is zero on UHS.
( therefore 2 sin ^{2}left(frac{A-B}{2}right)=0 )
men ( A=B )
and when ( sin A sin B(1-sin C)=0 )
( 0

# In any triangle ABC, if cos A cos B + sin Asin B sin C=1 then prove that triangle is an isosceles right angled.

Solution