Question

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

Answer 1

Hint:First of all consider a triangle ABC with C as the angle \[\theta \]. Now, use the Pythagoras theorem to find the remaining side. Now, find \[\sin \theta =\dfrac{P}{H}\] and \[\tan \theta =\dfrac{P}{B}\] and substitute in the given expression to get the required answer.

Complete step-by-step answer:
Here, we are given that \[\cos \theta =\dfrac{5}{13}\]. We have to find the value of \[\dfrac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{2\sin \theta \cos \theta }\times \dfrac{1}{{{\tan }^{2}}\theta }\].
Let us consider the expression given in the question.
\[E=\dfrac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{2\sin \theta \cos \theta }\times \dfrac{1}{{{\tan }^{2}}\theta }....\left( i \right)\]
We are given that,
\[\cos \theta =\dfrac{5}{13}....\left( ii \right)\]
We know that,
\[\cos \theta =\dfrac{Base}{Hypotenuse}....\left( iii \right)\]
From equation (ii) and (iii), we get,
\[\dfrac{5}{13}=\dfrac{Base}{Hypotenuse}\]
Let us assume a triangle ABC, right-angled at B and angle C is \[\theta \].

Let base BC be equal to 5x and hypotenuse equal to 13x. We know that Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides. So, in the above triangle ABC, by applying Pythagoras theorem, we get,
\[{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}\]
By substituting the value of BC = 5x and AC = 13x, we get,
\[{{\left( 5x \right)}^{2}}+{{\left( AB \right)}^{2}}={{\left( 13x \right)}^{2}}\]
\[25{{x}^{2}}+{{\left( AB \right)}^{2}}=169{{x}^{2}}\]
\[{{\left( AB \right)}^{2}}=169{{x}^{2}}-25{{x}^{2}}\]
\[{{\left( AB \right)}^{2}}=144{{x}^{2}}\]
\[AB=12x\]
We know that,
\[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}....\left( iv \right)\]
In triangle ABC, with respect to angle \[\theta \],
Perpendicular = AB = 12x
Hypotenuse = AC = 13x
By substituting these values in equation (iv), we get,
\[\sin \theta =\dfrac{12x}{13x}=\dfrac{12}{13}\]
We also know that,
\[\tan \theta =\dfrac{Perpendicular}{Base}....\left( v \right)\]
In triangle ABC, with respect to angle \[\theta \],
Perpendicular = AB = 12x
Base = BC = 5x
By substituting these values in equation (v), we get,
\[\tan \theta =\dfrac{12x}{5x}=\dfrac{12}{5}\]
Now, by substituting these values of \[\cos \theta ,\sin \theta \] and \[\tan \theta \] in equation (i), we get,
\[E=\dfrac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{2\sin \theta \cos \theta }.\dfrac{1}{{{\tan }^{2}}\theta }\]
\[E=\dfrac{{{\left( \dfrac{12}{13} \right)}^{2}}-{{\left( \dfrac{5}{13} \right)}^{2}}}{2.\left( \dfrac{12}{13} \right).\left( \dfrac{5}{13} \right)}.\dfrac{1}{{{\left( \dfrac{12}{5} \right)}^{2}}}\]
\[E=\left[ \dfrac{\left( \dfrac{144}{169}-\dfrac{25}{169} \right)}{\dfrac{120}{169}} \right].\left( \dfrac{25}{144} \right)\]
\[E=\left[ \dfrac{\left( \dfrac{144-25}{169} \right)}{\dfrac{120}{169}} \right].\left( \dfrac{25}{144} \right)\]
\[E=\dfrac{119}{169}\times \dfrac{169}{120}\times \dfrac{25}{144}\]
\[E=\dfrac{119}{120}\times \dfrac{25}{144}\]
\[E=\dfrac{119\times 5}{24\times 144}\]
\[E=\dfrac{595}{2736}\]
So, we get the value of \[\dfrac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{2\sin \theta \cos \theta }\times \dfrac{1}{{{\tan }^{2}}\theta }\] as \[\dfrac{595}{2736}\].

Note: In this question, students can also find \[\sin \theta \] by using \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. After getting \[\sin \theta \], students can find \[\tan \theta \] by using \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Also, students must keep in mind that when nothing is given about the angle, we should always consider it in the first quadrant, i.e between 0 to \[\dfrac{\pi }{2}\] and in this case all trigonometric ratios would be positive.Students should remember important trigonometric ratios for solving these types of questions.