Trigonometric Identities Question 187
Question: If $ x{{\sin }^{3}}\alpha +y{{\cos }^{3}}\alpha =\sin \alpha \cos \alpha $ and $ x\sin \alpha -y\cos \alpha =0, $ then $ x^{2}+y^{2}= $
[WB JEE 1984]
Options:
A) - 1
B) ±1
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ x,{{\sin }^{3}}\alpha +y,{{\cos }^{3}}\alpha =\sin ,\alpha ,\cos ,\alpha $ ?..(i) and $ x,\sin ,\alpha -y,\cos ,\alpha =0 $ ?..(ii) Now from (ii), $ x,\sin ,\alpha =y,\cos ,\alpha $ Putting in (i), we get
$ \Rightarrow y,\cos \alpha ,{{\sin }^{2}}\alpha +y,{{\cos }^{3}}\alpha =\sin ,\alpha ,\cos ,\alpha $
$ \Rightarrow y,\cos \alpha ,{ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }=\sin ,\alpha ,\cos ,\alpha $
$ \Rightarrow y,\cos ,\alpha =\sin ,\alpha ,\cos ,\alpha ,\Rightarrow y=\sin ,\alpha $ and $ x=\cos ,\alpha $ Hence, $ x^{2}+y^{2}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1. $
BETA
