त्रिभुज ABC में, sin A - cos B = cos C तो कोण B ज्ञात करें:
Solution
$\sin A = \cos B + \cos C $
$2 \sin \dfrac{A}{2} \cos \dfrac{A}{2} = 2 \cos \left( \dfrac{B + C}{2} \right) \cos \left( \dfrac{B-C}{2}\right)$
$A + B + C = 180^{\circ}$
$\dfrac{B + C}{2} = \dfrac{180 -A}{2} \Rightarrow 90 - \dfrac{A}{2}$
$\sin \dfrac{A}{2} = \cos \left( 90 - \dfrac{A}{2}\right) \cos \left( \dfrac{B-C}{2}\right)$
$\Rightarrow \sin \dfrac{A}{2} \left[ \cos \dfrac{A}{2} - \cos \left( \dfrac{B-C}{2} \right) \right]=0 $
$\cos \dfrac{A}{2} - \cos \left( \dfrac{B-C}{2}\right) = 0$
$2 \sin \left( \dfrac{A + B - C}{4}\right) \sin \left( \dfrac{B-C-A}{4}\right) = 0$
$\sin \left( \dfrac{A + B -C}{4} \right) = 0$ या $\sin \left( \dfrac{B-C-A}{4}\right) = 0$
A + B - C = 0 या B - C - A = 0
A + B + C = $180^{\circ}$
A + C = 180 - B
$\Rightarrow 2B = 180 \Rightarrow B = 90^{\circ} = \dfrac{\pi}{2}$