Question

3 cot theta is equal to 2 find 4 sin theta minus 3 cos theta divided by 2 sin theta + 6 cos theta

Answer 1

Hello !

Given ,

3 cotθ = 2

so,
cotθ = 2/3

we know that ,
cot θ = adjacent side/opposite side
         = 2/3

If a , b and c are sides of the right triangle ,

a = 2k
b = 3k


hypotenuse = c = (2k)² + (3k)² = 13k² 
                    = √13k

Now ,
sinθ = opposite side/hypotenuse = 3k/√13k = 3/√13
cosθ = adjacent side/hypotenuse = 2k/√13k = 2/√13

Now lets substitute the given values in the qn :-

{4 sinθ - 3cosθ}/{2sinθ + 6cosθ}

{4*3 ÷ √13 - 3* 2 ÷ √13 }/{ 2 * 3 ÷√13 + 6* 2÷√13}

{12÷√13 - 6 ÷ √13 } / { 6÷√13 + 12÷√13}

{6÷√13}/ {18÷√13}

6 / √13 * √13 / 18

√13 gets cancelled,

6/18 = 1/3

The ans is 1/3

Answer 2

Answer:

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

Step-by-step explanation:

Given : $3\cot \theta=2$

To find : The value of $\frac{4\sin\theta-3\cos\theta}{2\sin\theta+6\cos\theta}$ ?

Solution :  

$3\cot \theta=2$

$\cot \theta=\frac{2}{3}$

The expression is $\frac{4\sin\theta-3\cos\theta}{2\sin\theta+6\cos\theta}$

Divide numerator and denominator by $\sin\theta$

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

$=\frac{4-3\cot\theta}{2+6\cot\theta}$

$=\frac{4-3(\frac{2}{3})}{2+6(\frac{2}{3})}$

$=\frac{4-2}{2+4}$

$=\frac{2}{6}$

$=\frac{1}{3}$

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