Question

If $x\sin \theta +y\cos \theta =2$ and $x\sin \theta -y\cos \theta =0$ then $\frac{1}{x^2}+\frac{1}{y^2}$ is

• A. greater than 0
• B. less than equal to 1
• C. equal to $(-1)$
• D. less than (-1)

Answer 1

Answer: (A), (B)

Given: $x\sin \theta -y\cos \theta =0$
$\Rightarrow x\sin \theta =y\cos \theta$
$\text{ and }x\sin \theta +y\cos \theta =2$
$\Rightarrow y\cos \theta +y\cos \theta =2\Rightarrow 2y\cos \theta =2$
$\Rightarrow y\cos \theta =1$
$\therefore x\sin \theta =1\text{ and }y\cos \theta =1$
$\sin \theta =\frac{1}{x}\cos \theta =\frac{1}{y}$
Squaring the above equations, we get
${\sin }^2\theta =\frac{1}{x^2}$...(i)
${\cos }^2\theta =\frac{1}{y^2}$...(ii)
Adding equations (i) and (ii)
${\sin }^2\theta +{\cos }^2\theta =\frac{1}{x^2}+\frac{1}{y^2}=1$