In A B C, 1+ (A-B) C 1+ (A-C) B = - Vidyadip

MCQ

In $\triangle A B C, \frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=$

A
$\frac{a-b}{a-c}$
B
$\frac{a+b}{a+c}$
C
$\frac{a^2-b^2}{a^2-c^2}$
✓
$\frac{a^2+b^2}{a^2+c^2}$

✓

Answer

Correct option: D.

$\frac{a^2+b^2}{a^2+c^2}$

(D) $\frac{1+\cos C \cos (A-B)}{1+\cos (A-C) \cos B}=\frac{1-\cos (A+B) \cos (A-B)}{1-\cos (A-C) \cos (A+C)}$
$=\frac{1-\frac{1}{2}(\cos 2 A+\cos 2 B)}{1-\frac{1}{2}(\cos 2 A+\cos 2 C )}$
$=\frac{1-\frac{1}{2}\left(1-2 \sin ^2 A+1-2 \sin ^2 B\right)}{1-\frac{1}{2}\left(1-2 \sin ^2 A+1-2 \sin ^2 C \right)}$
$=\frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A+\sin ^2 C }=\frac{ a ^2+ b ^2}{ a ^2+ c ^2}$

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