Question

If $\cos A = \dfrac{3}{5}$, evaluate $\dfrac{{5\sin A + 3\sec A - 3\tan A}}{{4\cot A + 4\cos {\text{ec}}A + 5\cos A}}$.

Answer 1

Express $\cos A = \dfrac{3}{5}$ as $\dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$ and then calculate the value of perpendicular using the Pythagoras theorem. Next, find all other five ratios using the formula of their ratios, like, $\sin A = \dfrac{P}{H}$. At last substitute the values of trigonometric ratios in the given expression to determine its value.

Complete step-by-step answer:
We are given that $\cos A = \dfrac{3}{5}$
Also, we know that $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$
Therefore, base is 3 and hypotenuse is 5. We can calculate the perpendicular using the Pythagoras theorem
${H^2} = {B^2} + {P^2}$
On substituting the values of base and hypotenuse, we will get,
${\left( 5 \right)^2} = {\left( 3 \right)^2} + {P^2} \\\ {P^2} = 25 - 9 \\\ {P^2} = 16 \\\ P = 4 \\\$
Now, we will find all other 5 trigonometric ratios,
$\sin A = \dfrac{P}{H} = \dfrac{4}{5}$
$\sec A = \dfrac{H}{B} = \dfrac{5}{3}$
${\text{cosec}}A = \dfrac{H}{P} = \dfrac{5}{4}$
$\tan A = \dfrac{P}{B} = \dfrac{4}{3}$
$\cot A = \dfrac{B}{P} = \dfrac{3}{4}$
Now, substitute the values in the given expression to find its value
=$\dfrac{{5\left( {\dfrac{4}{5}} \right) + 3\left( {\dfrac{5}{3}} \right) - 3\left( {\dfrac{4}{3}} \right)}}{{4\left( {\dfrac{3}{4}} \right) + 4\left( {\dfrac{5}{4}} \right) + 5\left( {\dfrac{3}{5}} \right)}}$
We will simplify the expression,
$\Rightarrow$ $\dfrac{{4 + 5 - 4}}{{3 + 5 + 3}} = \dfrac{5}{{11}}$

Note: We can also calculate the values of cosecant is reciprocal of sine ratio and secant is reciprocal of cosine ratio. Also, tangent of any angle can be calculated by dividing sine ratio by cosine ratio. We have used Pythagoras theorem to find the perpendicular and hence the value of sine of the angle but we can also use the identity ${\cos ^2}x + {\sin ^2}x = 1$ to find the value of sine.