CBSE BoardEnglish MediumSTD 10MathsTrigonometric Ratios5 Marks
Question
If $3\cot\theta=2,$ find the value of $\frac{4\sin\theta-3\cos\theta}{2\sin\theta+6\cos\theta}.$
✓
Answer
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,
$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{1}{3}$
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