Question

If $\cos \theta =\dfrac{3}{5}$, find the value of $\left( 5\cos ec\theta -4\tan \theta \right)\left( \sec \theta +\cot \theta \right)$.

Answer 1

Hint:Assume that in the given function: $\cos \theta =\dfrac{3}{5}$, 3 is the length of base and 5 is the length of hypotenuse of a right angle triangle. Use Pythagoras theorem given by: $\text{hypotenuse}{{\text{e}}^{\text{2}}}=\text{bas}{{\text{e}}^{\text{2}}}+\text{perpendicular}{{\text{r}}^{\text{2}}}$, to determine the length of the perpendicular of the right angle triangle. Now, find $\sin \theta$ by taking the ratio of perpendicular and hypotenuse. Now, simplify the expression: $\left( 5\cos ec\theta -4\tan \theta \right)\left( \sec \theta +\cot \theta \right)$ by changing all the trigonometric functions into its sine and cosine form. Finally, substitute the value of $\sin \theta$ and $\cos \theta$ in the expression: $\left( 5\cos ec\theta -4\tan \theta \right)\left( \sec \theta +\cot \theta \right)$ to get the answer.

Complete step-by-step answer:
We have been provided with the trigonometric ratio, $\cos \theta =\dfrac{3}{5}$.
We know that, $\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}$. Therefore, on comparing it with the above provided ratio, we have, 3 as the length of perpendicular and 5 as the length of hypotenuse of a right angle triangle.

Now, using Pythagoras theorem: $\text{hypotenuse}{{\text{e}}^{\text{2}}}=\text{bas}{{\text{e}}^{\text{2}}}+\text{perpendicular}{{\text{r}}^{\text{2}}}$, we get,

& \text{perpendicular}{{\text{r}}^{\text{2}}}=\text{hypotenuse}{{\text{e}}^{\text{2}}}-\text{bas}{{\text{e}}^{\text{2}}} \\\ & \Rightarrow \text{perpendicular}=\sqrt{\text{hypotenuse}{{\text{e}}^{\text{2}}}-\text{bas}{{\text{e}}^{\text{2}}}} \\\ & \Rightarrow \text{perpendicular}=\sqrt{{{5}^{2}}-{{3}^{\text{2}}}} \\\ & \Rightarrow \text{perpendicular}=\sqrt{25-9} \\\ & \Rightarrow \text{perpendicular}=\sqrt{16} \\\ & \Rightarrow \text{perpendicular}=4 \\\ \end{aligned}$$ We know that, $$\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$$. $\Rightarrow \sin \theta =\dfrac{4}{5}$ Here, we have to find the value of the expression: $$\left( 5\cos ec\theta -4\tan \theta \right)\left( \sec \theta +\cot \theta \right)$$. First, let us simplify this expression by changing all the trigonometric functions into its sine and cosine form. Therefore, $$\begin{aligned} & \left( 5\cos ec\theta -4\tan \theta \right)\left( \sec \theta +\cot \theta \right) \\\ & =\left( \dfrac{5}{\sin \theta }-\dfrac{4\sin \theta }{\cos \theta } \right)\left( \dfrac{1}{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \right) \\\ \end{aligned}$$ Now, substituting $\cos \theta =\dfrac{3}{5}$ and $\sin \theta =\dfrac{4}{5}$ in the above simplified form, we get, $$\begin{aligned} & \left( 5\cos ec\theta -4\tan \theta \right)\left( \sec \theta +\cot \theta \right) \\\ & =\left( \dfrac{5}{\sin \theta }-\dfrac{4\sin \theta }{\cos \theta } \right)\left( \dfrac{1}{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \right) \\\ & =\left( \dfrac{5}{\dfrac{4}{5}}-\dfrac{4\times \dfrac{4}{5}}{\dfrac{3}{5}} \right)\left( \dfrac{1}{\dfrac{3}{5}}+\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} \right) \\\ & =\left( \dfrac{25}{4}-\dfrac{16}{3} \right)\left( \dfrac{5}{3}+\dfrac{3}{4} \right) \\\ \end{aligned}$$ Taking L.C.M we get, $\begin{aligned} & =\left( \dfrac{75-64}{12} \right)\left( \dfrac{20+9}{12} \right) \\\ & =\dfrac{11}{12}\times \dfrac{29}{12} \\\ & =\dfrac{319}{144} \\\ \end{aligned}$ Note: You may also solve the question without converting all the trigonometric functions into sine and cosine form. But then you have to find all the trigonometric ratios one by one. We have used the conversion, so that we have to find only sine of the given angle. It is important to note that we are dealing with the trigonometry of angles of the first quadrant only, therefore, all the values taken are positive.