• # question_answer If $\mathbf{ta}{{\mathbf{n}}^{\mathbf{2}}}\phi =\mathbf{2ta}{{\mathbf{n}}^{\mathbf{2}}}\phi +1$,then $cos2\phi +si{{n}^{2}}\phi$is equal to A)  0                                 B)  $-1$ C)  1                                 D)  2

[a] $\cos 2\phi +{{\sin }^{2}}\phi =\frac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi }+{{\sin }^{2}}\phi$ $=\frac{1-2{{\tan }^{2}}\phi -1}{1+2{{\tan }^{2}}\phi +1}+{{\sin }^{2}}\phi =\frac{-2{{\tan }^{2}}\phi }{2(1+{{\tan }^{2}}\phi )}+{{\sin }^{2}}\phi$ $=\frac{\frac{{{\sin }^{2}}\phi }{{{\cos }^{2}}\phi }}{1+\frac{{{\sin }^{2}}\phi }{{{\cos }^{2}}\phi }}+{{\sin }^{2}}\phi =\frac{-{{\sin }^{2}}\phi }{{{\cos }^{2}}\phi +{{\sin }^{2}}\phi }+{{\sin }^{2}}\phi$ $[\because {{\cos }^{2}}\phi +{{\sin }^{2}}\phi =1]$ $=-si{{n}^{2}}\phi +si{{n}^{2}}\phi$ = 0