Question

If $asin\theta +bcos\theta =acosec\theta -bsec\theta =1,$ prove that $a^2+b^2=1+b^{2/3}-b^{4/3}$

Answer 1

$asin\theta +bcos\theta =1$
$acos\theta -bsin\theta =sin\theta \cdot cos\theta$ square and add
$a^2+b^2=1+si{n}^2\theta co{s}^2\theta$
$=1+co{s}^2\theta -co{s}^4\theta$
Now we have to get only b in L. H. S.
Multiply the given equation by $cos\theta andsin\theta$ and subtracting , we get
$b(co{s}^2\theta +si{n}^2\theta )=cos\theta -cos\theta \cdot si{n}^2\theta$
or b = $co{s}^3\theta \therefore cos\theta =b^{1/3}$
$\therefore a^2+b^2=1+b^{2/3}-b^{4/3}$