Question

If $x=\csc \theta -\sin \theta$ and $y=\sec \theta -\cos \theta$, then the value of $x^2+y^2$ $=cose{c}^2\theta .se{c}^2\theta -p$ . Then $p=$

Answer 1

Given:

$x=cosec\theta -sin\theta$ and

$y=sec\theta -cos\theta$

Hence,

$x^2+y^2=cose{c}^2\theta +si{n}^2\theta -2cosec\theta sin\theta +se{c}^2\theta +co{s}^2\theta -2sec\theta cos\theta$

$=1+1+ta{n}^2\theta +1+co{t}^2\theta -2-2=\frac{si{n}^2\theta }{co{s}^2\theta }+\frac{co{s}^2\theta }{si{n}^2\theta }-1=cose{c}^2\theta se{c}^2\theta -1$...................[since, $si{n}^{2}A+co{s}^{2}A=1;1+ta{n}^{2}A=se{c}^{2}A;1+co{t}^{2}A=cose{c}^{2}A]$

Hence,

$P=1$