Trigonometric Identities Question 183
Question: If $ \theta $ is eliminated from the equations $ x=a,\cos (\theta -\alpha ) $ and $ y=b,\cos (\theta -\beta ) $ , then $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{2xy}{ab}\cos (\alpha -\beta ) $ is equal to
Options:
A) $ {{\sec }^{2}}(\alpha -\beta ) $
B) $ \cos ec^{2}(\alpha -\beta ) $
C) $ {{\cos }^{2}}(-\beta ) $
D) $ sin^{2}(\alpha -\beta ) $
Show Answer
Answer:
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 {{[ \frac{xy}{ab}-\cos (\alpha -\beta ) ]}^{2}}=( 1-\frac{x^{2}}{a^{2}} )( 1-\frac{y^{2}}{b^{2}} ) $ 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 )=sin^{2}(\alpha -\beta ) $
BETA
