Question

If $ x\,\,\sin \theta -y\,\cos \,\theta =0 $ and $ x\,{{\sin }^{3}}\theta +y\,{{\cos }^{3}}\theta =\sin \theta \,\,\cos \theta , $ then $ {{x}^{2}}+{{y}^{2}} $ is equal to

• A. $ 0 $
• B. $ 2 $
• C. $ 4 $
• D. $ 1 $

Answer 1

Answer: (D)

Given, $ x\,\sin \theta -y\,\cos \,\theta =0 $ ..(i)
and $ x\,{{\sin }^{3}}\theta +y{{\cos }^{3}}\theta =\sin \theta \,\cos \theta $ ..(ii)
On multiplying Eq. (i) by $ {{\sin }^{2}}\theta $
and subtracting from Eq. (ii), we get
$ y\,{{\cos }^{3}}\theta +y\cos \,{{\sin }^{2}}\theta =\sin \theta \,\cos \theta $
$ \Rightarrow $ $ y\,\cos \theta (1)=sin\theta \,cos\theta $
$ \Rightarrow $ $ y=\sin \theta $
From Eq. (i), $ x\,\sin \theta =\sin \theta \,\cos \theta $
$ \Rightarrow $ $ x=\cos \theta $
Now, $ {{x}^{2}}+{{y}^{2}}={{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 $