Question

If $3sin\theta +5cos\theta =5,$ then find the value of $5sin\theta -3cos\theta .$

Answer 1

We have, $3sin\theta +5cos\theta =5$
Square both the sides we get,
$9\left(sin\right)^{2}\theta +25\left(cos\right)^{2}\theta +30sin\theta cos\theta =25...\left(i\right)$
Let, $5sin\theta -3cos\theta =x$
Square both the sides we get,
$25\left(sin\right)^{2}\theta +9\left(cos\right)^{2}\theta -30sin\theta cos\theta =x^{2}...\left(i i\right)$
Adding equation $\left(i\right)\&\left(i i\right)$ we get,
$25\left(\left(sin\right)^{2} \theta + \left(cos\right)^{2} \theta \right)+9\left(\left(sin\right)^{2} \theta + \left(cos\right)^{2} \theta \right)=25+x^{2}$
$\Rightarrow 9+25=x^{2}+25$
$\Rightarrow x^{2}=9$
$\Rightarrow x=\pm3$
$\therefore x=3$