Trigonometric Identities Question 215
Question: If $ \tan \theta -\cot \theta =a $ and $ \sin \theta +\cos \theta =b, $ then $ {{(b^{2}-1)}^{2}}(a^{2}+4) $ is equal to
[WB JEE 1979]
Options:
A) 2
B) - 4
C) ± 4
D) 4
Show Answer
Answer:
Correct Answer: D
Solution:
Given that $ \tan \theta -\cot \theta =a $ ?..(i) and $ \sin \theta +\cos \theta =b $ ?..(ii) Now $ {{(b^{2}-1)}^{2}}(a^{2}+4) $ $ ={{{ {{(\sin \theta +\cos \theta )}^{2}}-1 }}^{2}}{ {{(\tan \theta -\cot \theta )}^{2}}+4 } $ $ ={{[1+\sin 2\theta -1]}^{2}}[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta -2+4] $ $ ={{\sin }^{2}}2\theta (cose{c^{2}}\theta +{{\sec }^{2}}\theta ) $ $ =4{{\sin }^{2}}\theta {{\cos }^{2}}\theta [ \frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta } ]=4 $ . Trick: Obviously the value of expression $ {{(b^{2}-1)}^{2}}(a^{2}+4) $ is independent of $ \theta $ , therefore put any suitable value of $ \theta $ . Let $ \theta =45{}^\circ $ , we get $ a=0,\ b=\sqrt{2} $ so that $ {{[{{(\sqrt{2})}^{2}}-1]}^{2}} $ $ (0^{2}+4)=4. $
BETA
