Question

If 4 sin^2theta+9 cos^2 theta=12 sin theta cos theta.Then find 4sin theta+3 cos theta/5 sin theta -4cos theta​

Answer 1

Given:

$4sin^{2}\theta+9cos^{2}\theta=12sin\theta cos\theta$

To find:

$\dfrac{4sin\theta+3cos\theta}{5sin\theta-4cos\theta}$

Step-by-step explanation:

Now, $4sin^{2}\theta+9cos^{2}\theta=12sin\theta cos\theta$

$\Rightarrow 4sin^{2}\theta-12sin\theta cos\theta+9cos^{2}\theta=0$

$\Rightarrow (2sin\theta-3cos\theta)^{2}=0$

$\Rightarrow 2sin\theta-3cos\theta=0$

$\Rightarrow 2sin\theta=3cos\theta$

$\Rightarrow \dfrac{sin\theta}{cos\theta}=\dfrac{3}{2}$

$\therefore \dfrac{4sin\theta+3cos\theta}{5sin\theta-4cos\theta}$

• divide both the numerator and the denominator by $cos\theta\neq 0$

$=\dfrac{4\dfrac{sin\theta}{cos\theta}+3}{5\dfrac{sin\theta}{cos\theta}-4}$

$=\dfrac{4\times\dfrac{3}{2}+3}{5\times\dfrac{3}{2}-4}$

• since $\dfrac{sin\theta}{cos\theta}=\dfrac{3}{2}$

$=\dfrac{6+3}{\dfrac{15-8}{2}}$

$=\dfrac{2\times 9}{7}$

$=\dfrac{18}{7}$

Answer:

$\dfrac{4sin\theta+3cos\theta}{5sin\theta-4cos\theta}=\dfrac{18}{7}$