If \[\mathbf{tan}\alpha =\mathbf{ytan}\beta \] and \[\mathbf{sin}\alpha =\mathbf{xsin}\beta \], then \[\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}\alpha \] is

\[\frac{{{x}^{2}}+1}{{{y}^{2}}+1}\]

\[\frac{{{x}^{2}}}{{{y}^{2}}+1}\]

\[\frac{{{x}^{2}}}{{{y}^{2}}}\]

\[\frac{{{x}^{2}}-1}{{{y}^{2}}-1}\]

If \[\mathbf{tan}\alpha =\mathbf{ytan}\beta \] and \[\mathbf{sin}\alph

1.	If \[\mathbf{tan}\alpha =\mathbf{ytan}\beta \] and \[\mathbf{sin}\alpha =\mathbf{xsin}\beta \], then \[\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}\alpha \] is
A.	\[\frac{{{x}^{2}}}{{{y}^{2}}+1}\]
B.	\[\frac{{{x}^{2}}}{{{y}^{2}}}\]
C.	\[\frac{{{x}^{2}}-1}{{{y}^{2}}-1}\]
D.	\[\frac{{{x}^{2}}+1}{{{y}^{2}}+1}\]
Answer» D. \[\frac{{{x}^{2}}+1}{{{y}^{2}}+1}\]

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If \[\mathbf{tan}\alpha =\mathbf{ytan}\beta \] and \[\mathbf{sin}\alpha =\mathbf{xsin}\beta \], then \[\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}\alpha \] is

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