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# If $\cos \theta =\dfrac{5}{13}$ , find the value of $\dfrac{2\sin \theta -{{\cos }^{2}}\theta }{2\sin \theta \cos \theta }\times \dfrac{1}{{{\tan }^{2}}\theta }$ . 

Last updated date: 24th Jun 2024
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Hint:
We recall the definitions of sine, cosine and tangent trigonometric ratios from the right angled triangle. We take the length of the adjacent side as $b=5$ and hypotenuse as $h=13$ . We find the side $b$ using Pythagoras theorem. We find $\sin \theta =\dfrac{p}{h},\tan \theta =\dfrac{p}{b}$ and put the values in given trigonometric expression.

We know that in right angled triangle (here $\Delta ABC$ ) the side opposite to right angled triangle is called hypotenuse denoted as $h=AC$ , the vertical side is called perpendicular denoted as $p=AB$ and the horizontal side is called the base denoted as $b=BC$ .

We know from the trigonometric ratios in a right angled triangle the sine of any angle is given by the ratio of side opposite to the angle to the hypotenuse. In the figure the sine of the angle $\theta$ is given by
$\sin \theta =\dfrac{p}{h}$
Similarly the cosine of an angle is the ratio of side adjacent to the angle (excluding hypotenuse) to the hypotenuse. So we have cosine of angle $\theta$
$\cos \theta =\dfrac{b}{h}.$
The tangent of the angle is the ratio of opposite side to the adjacent side (excluding hypotenuse) . So we have tangent of the angle of angle $\theta$
$\tan \theta =\dfrac{p}{b}$
We know from Pythagoras theorem that “in a right-angled triangle the square of the hypotenuse is the sum of squares of the other two sides”. So in triangle we have
\begin{align} & {{h}^{2}}={{p}^{2}}+{{b}^{2}} \\ & \Rightarrow {{p}^{2}}={{h}^{2}}-{{b}^{2}} \\ & \Rightarrow p=\sqrt{{{h}^{2}}-{{b}^{2}}} \\ \end{align}
We are give in the question that $\cos \theta =\dfrac{5}{13}$ . So we have;
$\cos \theta =\dfrac{5}{13}=\dfrac{b}{h}$
So let us take $b=5,h=13$ . So we have
$p=\sqrt{{{13}^{2}}-{{5}^{2}}}=\sqrt{169-25}=\sqrt{144}=12$
So we have sine and tangent of the angle as;
\begin{align} & \sin \theta =\dfrac{p}{h}=\dfrac{12}{13} \\ & \tan \theta =\dfrac{p}{b}=\dfrac{12}{5} \\ \end{align}
We pout the obtained values of $\sin \theta =\dfrac{12}{13},\tan \theta =\dfrac{12}{5}$ and give value of $\cos \theta =\dfrac{5}{13}$ in the given expression to have;
\begin{align} & \dfrac{2\sin \theta -{{\cos }^{2}}\theta }{2\sin \theta \cos \theta }\times \dfrac{1}{{{\tan }^{2}}\theta } \\ & \Rightarrow \dfrac{2\times \dfrac{12}{13}-\dfrac{{{5}^{2}}}{{{13}^{2}}}}{2\times \dfrac{12}{13}\times \dfrac{5}{13}}\times \dfrac{1}{{{\left( \dfrac{12}{5} \right)}^{2}}} \\ & \Rightarrow \dfrac{\dfrac{2\times 12\times 13-{{5}^{2}}}{{{13}^{2}}}}{\dfrac{2\times 12\times 5}{{{13}^{2}}}}\times \dfrac{{{5}^{2}}}{{{12}^{2}}} \\ \end{align}
We cancel out ${{13}^{2}}$ from the numerator and denominator to have;
\begin{align} & \Rightarrow \dfrac{312-25}{120}\times \dfrac{25}{144} \\ & \Rightarrow \dfrac{287}{24}\times \dfrac{5}{144}=\dfrac{1435}{3456} \\ \end{align}
So the evaluated value is $\dfrac{1435}{3456}$ .
Note:
We can alternatively solve by using Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find $\sin \theta$ and the $\tan \theta$ from $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . We note that lengths are always positive and that we have taken positive square roots only. The reciprocal ratios of sine, cosine and tangent are given as $\operatorname{cosec}\theta =\dfrac{h}{p},\sec \theta =\dfrac{h}{b},\cot \theta =\dfrac{b}{p}$ .