Trigonometric Identities Question 116
Question: If $ \cos ,(\theta -\alpha )=a,\sin ,(\theta -\beta )=b, $ then $ {{\cos }^{2}}(\alpha -\beta )+2ab,\sin ,(\alpha -\beta ) $ is equal to
Options:
A) $ 4a^{2}b^{2} $
B) $ a^{2}-b^{2} $
C) $ a^{2}+b^{2} $
D) $ -a^{2}b^{2} $
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Answer:
Correct Answer: C
Solution:
We have $ \sin (\alpha -\beta )=\sin (\theta -\beta -\overline{\theta -\alpha }) $ $ =\sin (\theta -\beta )\cos (\theta -\alpha )-\cos (\theta -\beta )\sin (\theta -\alpha ) $ $ =ba-\sqrt{1-b^{2}}\sqrt{1-a^{2}} $ and $ \cos (\alpha -\beta )=\cos (\theta -\beta -\overline{\theta -\alpha }) $ $ =\cos (\theta -\beta )\cos (\theta -\alpha )+\sin (\theta -\beta )\sin (\theta -\alpha ) $ $ =a\sqrt{1-b^{2}}+b\sqrt{1-a^{2}} $
$ \therefore $ Given expression is $ {{\cos }^{2}}(\alpha -\beta )+2ab\sin (\alpha -\beta ) $ $ =(a\sqrt{1-b^{2}}+b\sqrt{1-a^{2}{{)}^{2}}}+2ab{ab-\sqrt{1-a^{2}}\sqrt{1-b^{2}}} $ $ =a^{2}+b^{2} $ . Trick : Put $ \alpha =30{}^\circ ,\beta =60{}^\circ $ and $ \theta =90{}^\circ , $ then $ a=\frac{1}{2},b=\frac{1}{2} $
$ \therefore {{\cos }^{2}}(\alpha -\beta )+2ab\sin (\alpha -\beta )=\frac{3}{4}+\frac{1}{2}\times ( -\frac{1}{2} )=\frac{1}{2} $ which is given by option (c).
BETA
