Question

If the ratio of trigonometric ratio $\sin A:\cos A = 4:7$, then the value of $\dfrac{{7\sin A - 3\cos A}}{{7\sin A + 2\cos A}}$ is
$
  (a){\text{ }}\dfrac{3}{{14}} \\
  (b){\text{ }}\dfrac{3}{2} \\
  (c){\text{ }}\dfrac{1}{3} \\
  (d){\text{ }}\dfrac{1}{6} \\
 $

Answer 1

Hint: In this question take cosA common from both the numerator and denominator part. Then use the given ratio to simplify the expression and get its value.

Complete step-by-step answer:
Given data

$\sin A:\cos A = 4:7$

$ \Rightarrow \dfrac{{\sin A}}{{\cos A}} = \dfrac{4}{7}$ ...................... (1)

Now we have to find out the value of $\dfrac{{7\sin A - 3\cos A}}{{7\sin A + 2\cos A}}$.

So take cos (A) common from numerator and denominator we have,

$ \Rightarrow \dfrac{{7\sin A - 3\cos A}}{{7\sin A + 2\cos A}} = \dfrac{{\cos A\left(

{7\dfrac{{\sin A}}{{\cos A}} - 3} \right)}}{{\cos A\left( {7\dfrac{{\sin A}}{{\cos A}} + 2} \right)}}$

So as we see cos (A) is cancel out from numerator and denominator so we have,

$ \Rightarrow \dfrac{{7\sin A - 3\cos A}}{{7\sin A + 2\cos A}} = \dfrac{{\left( {7\dfrac{{\sin

A}}{{\cos A}} - 3} \right)}}{{\left( {7\dfrac{{\sin A}}{{\cos A}} + 2} \right)}}$

Now from equation (1) we have,

$ \Rightarrow \dfrac{{7\sin A - 3\cos A}}{{7\sin A + 2\cos A}} = \dfrac{{\left( {7 \times

\dfrac{4}{7} - 3} \right)}}{{\left( {7 \times \dfrac{4}{7} + 2} \right)}}$

Now simplify the above equation we have,

$ \Rightarrow \dfrac{{7\sin A - 3\cos A}}{{7\sin A + 2\cos A}} = \dfrac{{\left( {4 - 3}

\right)}}{{\left( {4 + 2} \right)}} = \dfrac{1}{6}$

So this is the required value.

Hence option (D) is correct.

Note: There could have been another way to solve this problem $\sin A:\cos A = \tan A$, using tanA the values of sinA and cosA could be taken out as $\tan A = \dfrac{P}{B}$, where P is the perpendicular and B is the base, so hypotenuse could be evaluated and thus sinA and cosA.