Trigonometric Identities Question 298
Question: If $ a\tan \theta =b $ , then $ a\cos 2\theta +b\sin 2\theta = $
[EAMCET 1981, 82; MP PET 1996; J & K 2005]
Options:
A) $ a $
B) $ b $
C) $ -a $
D) $ -b $
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ \tan \theta =\frac{b}{a} $ . Now, $ a\cos 2\theta +b\sin 2\theta =a( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } )+b( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } ) $ Putting $ \tan \theta =\frac{b}{a} $ , we get $ =a( \frac{1-\frac{b^{2}}{a^{2}}}{1+\frac{b^{2}}{a^{2}}} )+b( \frac{2\frac{b}{a}}{1+\frac{b^{2}}{a^{2}}} )=a( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} )+b( \frac{2ba}{a^{2}+b^{2}} ) $ $ =\frac{1}{(a^{2}+b^{2})}{a^{3}-ab^{2}+2ab^{2}}=\frac{a(a^{2}+b^{2})}{a^{2}+b^{2}}=a $ .
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