Trigonometric Ratios — Maths STD 10 — Question

Given: $3\cot\theta=2$

Therefore,

$\cot\theta=\frac{2}{3}\ \dots(1)$

Now, we know that $\cot\theta=\frac{\cos\theta}{\sin\theta}$

Therefore equuation (i) becomes

$\frac{\cos\theta}{\sin\theta}=\frac{2}{3}\ \dots(2)$

Now by applying Invetendo to equation (ii)

We get,

$\frac{\sin\theta}{\cos\theta}=\frac{3}{2}\ \dots(3)$

Now, multipalying by $\frac{4}{3}$ on botha sides

We get,

$\frac{4}{3}\times\frac{\sin\theta}{\cos\theta}=\frac{4}{3}\times\frac{3}{2}$

Therefore, 3 cancels out on R.H.S and

We get,

$\frac{4\sin\theta}{3\cos\theta}=\frac{2}{1}$

Now by applying dividendo in above equation

We get,

$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}=\frac{2-1}{1}$

$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}=\frac{1}{1}\ \dots(4)$

Now, multiplying by $\frac{2}{6}$ on both sides of equation (3)

We get,

$\frac{2}{6}\times\frac{\sin\theta}{\cos\theta}=\frac{2}{6}\times\frac{3}{2}$

Therefore, 2 cancels out on R.H.S and

We get,

$\frac{2\sin\theta}{6\cos\theta}=\frac{3}{6}$

$\frac{2\sin\theta}{6\cos\theta}=\frac{1}{2}$

Now by applying componendo in above equation

We get,

$\frac{2\cos\theta+6\sin\theta}{6\sin\theta}=\frac{1+2}{2}$

$\frac{2\cos\theta+6\sin\theta}{6\sin\theta}=\frac{3}{2}\ \dots(5)$

Now, by dividing equation (4) by equation (5)

We get,

$\frac{\frac{4\sin\theta-3\cos\theta}{3\sin\theta}}{\frac{2\cos\theta+6\sin\theta}{6\sin\theta}}=\frac{\frac{1}{1}}{\frac{3}{2}}$

Therefore,

$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}\times\frac{6\sin\theta}{2\cos\theta+6\sin\theta}=\frac{1}{1}\times\frac{2}{3}$

$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}\times\frac{2\times(3\sin\theta)}{2\cos\theta+6\sin\theta}=\frac{1}{1}\times\frac{2}{3}$

Therefore, on L.H.S $(3\sin\theta)$ cancels out and we get,

$\frac{2\times(4\sin\theta-3\cos\theta)}{2\cos\theta+6\sin\theta}=\frac{2}{3}$

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{2}{3\times2}$

Therefore, 2 cancels out on R.H.S. and

We get,

$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{1}{3}$

Hence,

$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{1}{3}$