Trigonometric Identities Question 297
Question: If $ \cos 2B=\frac{\cos (A+C)}{\cos (A-C)} $ , then $ \tan A,\ \tan B,\ \tan C $ are in
Options:
A) A.P.
B) G.P.
C) H.P.
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \cos 2B=\frac{\cos (A+C)}{\cos (A-C)}=\frac{\cos A\cos C-\sin A\sin C}{\cos A\cos C+\sin A\sin C} $
Þ $ \frac{1-{{\tan }^{2}}B}{1+{{\tan }^{2}}B}=\frac{1-\tan A\tan C}{1+\tan A\tan C} $
Þ $ 1+{{\tan }^{2}}B-\tan A\tan C-\tan A\tan C{{\tan }^{2}}B $ $ =1-{{\tan }^{2}}B+\tan A\tan C-\tan A\tan C{{\tan }^{2}}B $
Þ $ 2{{\tan }^{2}}B=2\tan A\tan C\Rightarrow {{\tan }^{2}}B=\tan A\tan C $ Hence, tan A, tan B and $ \tan $ C will be in G.P.
BETA
