Trigonometric Equations Question 240
Question: In triangle $ ABC, $ $ \frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}= $
Options:
A) $ \frac{a-b}{a-c} $
B) $ \frac{a+b}{a+c} $
C) $ \frac{a^{2}-b^{2}}{a^{2}-c^{2}} $
D) $ \frac{a^{2}+b^{2}}{a^{2}+c^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
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$ \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)} $
Þ $ \frac{1-{{\cos }^{2}}A+{{\sin }^{2}}B}{1-{{\cos }^{2}}A+{{\sin }^{2}}C}=\frac{{{\sin }^{2}}A+{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}C}=\frac{a^{2}+b^{2}}{a^{2}+c^{2}} $ .
BETA
