Question

If $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$ and $\sin 2\alpha =2\sin \alpha \cos \alpha$ then $(x\tan \alpha +y\cot \alpha )(x\cot \alpha +y\tan \alpha )-2xy{\cot }^{2}2\alpha$

• A. is equal to $(x+y{)}^2$
• B. is independent of $\alpha$
• C. is equal to $x\cos \alpha +y\sin \alpha$
• D. is independent of $x$ and $y$

Answer 1

Answer: (A), (B)

The correct options are
A is equal to $(x+y{)}^2$
B is independent of $\alpha$

$(x\tan \alpha +y\cot \alpha )(x\cot \alpha +y\tan \alpha )-2xy{\cot }^{2}2\alpha =(\frac{x\sin \alpha }{\cos \alpha }+\frac{y\cos \alpha }{\sin \alpha })(\frac{x\cos \alpha }{\sin \alpha }+\frac{y\sin \alpha }{\cos \alpha })-2xy\frac{{\cos }^{2}2\alpha }{{\sin }^{2}2\alpha }=\frac{(x{\sin }^2\alpha +y{\cos }^2\alpha )(x{\cos }^2\alpha +y{\sin }^2\alpha )}{{\sin }^2\alpha {\cos }^2\alpha }-\frac{2xy{({\cos }^2\alpha -{\sin }^2\alpha )}^2}{2{\sin }^2\alpha {\cos }^2\alpha }=\frac{x^2{\sin }^2\alpha {\cos }^2\alpha +y^2{\sin }^2\alpha {\cos }^2\alpha +xy{\cos }^4\alpha +xy{\sin }^4\alpha -xy{({\cos }^2\alpha -{\sin }^2\alpha )}^2}{{\sin }^2\alpha {\cos }^2\alpha }=\frac{{\sin }^2\alpha {\cos }^2\alpha (x^2+y^2)+2xy{\sin }^2\alpha {\cos }^2\alpha }{{\sin }^2\alpha {\cos }^2\alpha }=x^2+y^2+2xy={(x+y)}^2$