Question

If $x\cos \alpha +y\sin \alpha =x\cos \beta +y\sin \beta =a(0<\alpha ,\beta <\frac{\pi }{2})($, then the value of $\cos (\alpha +\beta )$ is

• A. $\frac{4axy}{x^2+y^2}$
• B. $\frac{2a^2-x^2-y^2}{x^2+y^2}$
• C. $\frac{x^2-y^2}{x^2+y^2}$
• D. $\frac{a^2-x^2}{a^2-y^2}$

Answer 1

Answer: (C)

$\frac{x^2-y^2}{x^2+y^2}$

$x\cos \theta +y\sin \theta =a$

() $y\sin \theta -a=-x\cos \theta$

squaring both sides,

$\Rightarrow (y\sin \theta -a{)}^2=(-x\cos \theta {)}^2$

$\Rightarrow y^2{\sin }^2\theta +a^2-2ay\sin \theta =x^2{\cos }^2\theta$

$\Rightarrow y^2{\sin }^2\theta +a^2-2ay\sin \theta =x^2(1-{\sin }^2\theta )$

( ${\sin }^2\theta +{\cos }^2\theta =1)$

$\Rightarrow {\sin }^2\theta (x^2+y^2)-2ay\sin \theta +(a^2-x^2)=0$

Which is a quadrotic equation

whose roots are $\sin \alpha$ and $\sin \beta$

$x\cos \theta +y\sin \theta =a$

$\Rightarrow x\cos \theta -a=-y\sin \theta$

squaring both sides,

$\Rightarrow (x\cos \theta -a{)}^2=(-y\sin \theta {)}^2$

$\Rightarrow x^2{\cos }^2\theta +a^2-2ax\cos \theta =y^2{\sin }^2\theta$

$\Rightarrow y^2({\cos }^2\theta -1)+x^2{\cos }^2\theta +a^2-2ax\cos \theta =0$

$({\sin }^2\theta +{\cos }^2\theta =1)$

$\Rightarrow {\cos }^2\theta (x^2+y^2)-2ax\cos \theta +(a^2-y^2)=0$

which is a quadratic equation

whose roots are $\cos \alpha$ and $\cos \beta$.

$\cos \alpha \cdot \cos \beta =\frac{a^2-y^2}{x^2+y^2}-\left(ii\right)$

$\begin{array}{cc}\text{Hence,}\cos (\alpha +\beta )& =\cos \alpha \cdot \cos \beta -\sin \alpha \sin \beta \\ & =\frac{a^2-y^2}{x^2+y^2}-\frac{a^2-x^2}{x^2+y^2}\\ & =\frac{x^2-y^2}{x^2+y^2}\end{array}$