Trigonometric Identities Question 241
Question: If $ \sin x+\sin y=a $ and $ cosx+cos,y=b, $ then $ {{\tan }^{2}}( \frac{x+y}{2} )+{{\tan }^{2}}( \frac{x-y}{2} ) $ is equal to
Options:
A) $ \frac{a^{4}+b^{4}+4b^{2}}{a^{2}b^{2}+b^{4}} $
B) $ \frac{a^{4}-b^{4}+4b^{2}}{a^{2}b^{2}+b^{4}} $
C) $ \frac{a^{4}-b^{4}+4a^{2}}{a^{2}b^{2}+a^{4}} $
D) None of the above
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Answer:
Correct Answer: B
Solution:
$ \sin x+\sin y=a $
$ \Rightarrow 2\sin ( \frac{x+y}{2} )\cos ( \frac{x-y}{2} )=a $ …(1) $ \cos x+\cos y=b $
$ \Rightarrow 2\cos ( \frac{x+y}{2} )\cos ( \frac{x-y}{2} )=b $ …(2) dividing eq. (1) & (2) $ \tan ( \frac{x+y}{2} )=\frac{a}{b} $ Squaring of eq. (1) & (2) and adding - $ 4{{\cos }^{2}}( \frac{x-y}{2} )=a^{2}+b^{2} $ $ sec^{2}( \frac{x-y}{2} )=\frac{4}{a^{2}+b^{2}} $ again- $ {{\tan }^{2}}( \frac{x+y}{2} )+{{\tan }^{2}}( \frac{x-y}{2} ) $ $ ={{( \frac{a}{b} )}^{2}}+{{\sec }^{2}}( \frac{x-y}{2} )-1 $ $ =\frac{a^{2}}{b^{2}}+\frac{4}{a^{2}+b^{2}}-1 $ $ =\frac{a^{4}-b^{4}+4b^{2}}{a^{2}b^{2}+b^{4}} $
BETA
