Question

If $\cot B = \dfrac{{12}}{5}$ , prove that ${\tan ^2}B - {\sin ^2}B = {\sin ^4}B{\sec ^2}B$

Answer 1

Hint: Use the identity, $\tan \theta = \dfrac{1}{{\cot \theta }}$and \[\tan \theta = \dfrac{P}{B}\]where P is perpendicular and B is base, to find perpendicular and base. Then use Pythagoras theorem ${H^2} = {P^2} + {B^2}$
Where H is the hypotenuse, P is the perpendicular and B is the base.
Then use the identities, $\sin \theta = \dfrac{P}{H}$and $\cos \theta = \dfrac{B}{H}$ and put the values in the equation to prove that LHS=RHS.

Complete step by step answer:

Given, $\cot B = \dfrac{{12}}{5}$
Now we know that $\tan \theta = \dfrac{1}{{\cot \theta }}$
So $\tan B = \dfrac{1}{{\dfrac{{12}}{5}}} = \dfrac{5}{{12}}$
Now, we also that \[\tan \theta = \dfrac{P}{B}\] , where P is perpendicular and B is the base of a triangle.
So P=$5$ and B=$12$
Then we have to find the Hypotenuse of the triangle.

According to Pythagoras theorem,
 In a right-angled triangle, the square of the longest side (hypotenuse) is equal to the sum of the squares of the two other sides. It is written as-
${H^2} = {P^2} + {B^2}$
Where H is the hypotenuse, P is the perpendicular and B is the base.
On putting the values in the formula we get,
$ \Rightarrow $ ${H^2} = {5^2} + {12^2}$
On solving we get,
$ \Rightarrow $ ${H^2} = 25 + 144$
On adding we get,
$ \Rightarrow $ ${H^2} = 169$
$ \Rightarrow H = \sqrt {169} = 13$
So we know that $\sin \theta = \dfrac{P}{H}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\cos \theta = \dfrac{B}{H}$
So $\sec \theta = \dfrac{H}{B}$
On putting values of P, H, and B we get,
$ \Rightarrow \sin B = \dfrac{5}{{13}}$ and $\sec B = \dfrac{{13}}{{12}}$
Now we have to prove ${\tan ^2}B - {\sin ^2}B = {\sin ^4}B{\sec ^2}B$
On taking LHS and putting the required values we get,
$ \Rightarrow {\tan ^2}B - {\sin ^2}B = {\left( {\dfrac{5}{{12}}} \right)^2} - {\left( {\dfrac{5}{{13}}} \right)^2}$
On taking $5$ common, we get-
$ \Rightarrow {\tan ^2}B - {\sin ^2}B = {5^2}\left[ {{{\left( {\dfrac{1}{{12}}} \right)}^2} - {{\left( {\dfrac{1}{{13}}} \right)}^2}} \right]$
On simplifying we get,
$ \Rightarrow {\tan ^2}B - {\sin ^2}B = 25\left[ {\dfrac{1}{{144}} - \dfrac{1}{{169}}} \right]$
On taking LCM we get,
$ \Rightarrow {\tan ^2}B - {\sin ^2}B = 25\left[ {\dfrac{{169 - 144}}{{144 \times 169}}} \right] = \dfrac{{25 \times 25}}{{169 \times 144}}$
On multiplying the numerator, we get
$ \Rightarrow {\tan ^2}B - {\sin ^2}B = \dfrac{{625}}{{144 \times 169}}$ --- (i)
On taking RHS and putting the required values we get,
\[ \Rightarrow {\sin ^4}B{\sec ^2}B = {\left( {\dfrac{5}{{13}}} \right)^4} \times {\left( {\dfrac{{13}}{{12}}} \right)^2}\]
On solving we get,
\[ \Rightarrow {\sin ^4}B{\sec ^2}B = \dfrac{{{5^4}}}{{{{13}^4}}} \times \dfrac{{{{13}^2}}}{{{{12}^2}}}\]
On cancelling ${13^2}$ from numerator and denominator, we get-
\[ \Rightarrow {\sin ^4}B{\sec ^2}B = \dfrac{{{5^4}}}{{{{13}^2} \times {{12}^2}}}\]
On simplifying we get,
\[ \Rightarrow {\sin ^4}B{\sec ^2}B = \dfrac{{625}}{{169 \times 144}}\] -- (ii)
From eq. (i) and eq. (ii), we get
$ \Rightarrow $ ${\tan ^2}B - {\sin ^2}B = {\sin ^4}B{\sec ^2}B$
Hence, Proved.

Note: You can also directly use $\cot \theta = \dfrac{B}{P}$where B=base and P=perpendicular. Then use the Pythagoras theorem to find hypotenuse (H). Also, you can use$\sec \theta = \dfrac{H}{B}$ to find the value of$\sec B$ .