A) \[{{\sec }^{2}}(\alpha -\beta )\]
B) \[\cos e{{c}^{2}}(\alpha -\beta )\]
C) \[{{\cos }^{2}}(-\beta )\]
D) \[si{{n}^{2}}(\alpha -\beta )\]
Correct Answer: D
Solution :
[d] \[(\alpha -\beta )=(\theta -\beta )-(\theta -\alpha )\] \[\Rightarrow \,\cos (\alpha -\beta )=cos(\theta -\beta )cos(\theta -\alpha)\] \[+\sin (\theta -\beta )sin(\theta -\alpha )\] \[=\frac{y}{b}\times \frac{x}{a}+\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}\sqrt{1-\frac{{{y}^{2}}}{{{b}^{2}}}}\] \[\Rightarrow {{\left[ \frac{xy}{ab}-\cos (\alpha -\beta ) \right]}^{2}}=\left( 1-\frac{{{x}^{2}}}{{{a}^{2}}} \right)\left( 1-\frac{{{y}^{2}}}{{{b}^{2}}} \right)\] or \[\frac{{{x}^{2}}{{y}^{2}}}{{{a}^{2}}{{b}^{2}}}+{{\cos }^{2}}(\alpha -\beta )-\frac{2xy}{ab}\cos (\alpha -\beta )\] \[=1-\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{x}^{2}}{{y}^{2}}}{{{a}^{2}}{{b}^{2}}}\] or \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{2xy}{ab}\cos (\alpha -\beta )=si{{n}^{2}}(\alpha -\beta )\]
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