Question

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

Answer 1

Answer: 3

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