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Question: If \(\cot \,B = \dfrac{{12}}{5}\), prove that \({\tan ^2}B - {\sin ^2}B = {\sin ^4}B{\sec ^2}B\)....

If cot⁡ B=125\cot \,B = \dfrac{{12}}{5}, prove that tan⁡2B−sin⁡2B=sin⁡4Bsec⁡2B{\tan ^2}B - {\sin ^2}B = {\sin ^4}B{\sec ^2}B.

Explanation

Solution

In this question we have to prove the left side of the equation equals the right side of the equation. We have cot⁡ B=125\cot \,B = \dfrac{{12}}{5}, by this value we can find the other trigonometric functions such as tan⁡B, sin⁡B, sec⁡B\tan B,\,\sin B,\,\sec B. We know that cot⁡\cot is the reciprocal of tan⁡\tan . Similarly, we can find sin⁡B\sin Band sec⁡B\sec B.

Complete step by step answer:
We have, cot⁡ B=125\cot \,B = \dfrac{{12}}{5}
We know that Cotθ=baseperpendicularCot\theta = \dfrac{\text{base}}{\text{perpendicular}}
As, tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }}
So, tan⁡B=512\tan B = \dfrac{5}{{12}}
Now, we have to find the value of sin⁡B\sin B and sec⁡B\sec B.
We know that
sin⁡θ=perpendicularhypotenuse\sin \theta = \dfrac{\text{perpendicular}}{\text{hypotenuse}} and sec⁡θ=hypotenusebase\sec \theta = \dfrac{\text{hypotenuse}}{\text{base}}
So, we have to find the hypotenuse.

In a right- angled triangle,
(hypotenuse)2=(perpendicular)2+(base)2{\text{(hypotenuse)}^2} = {\text{(perpendicular)}^2} + {\text{(base)}^2}
We have, perpendicular=5perpendicular = 5 and base=12base = 12
Putting the value of perpendicularperpendicular and basebase. We get,
⇒(hypotenuse)2=(5)2+(12)2\Rightarrow {\text{(hypotenuse)}^2} = {(5)^2} + {(12)^2}
⇒(hypotenuse)2=25+144\Rightarrow {\text{(hypotenuse)}^2} = 25 + 144
⇒(hypotenuse)2=169\Rightarrow {\text{(hypotenuse)}^2} = 169
⇒(hypotenuse)2=169\Rightarrow {\text{(hypotenuse)}^2} = \sqrt {169}
⇒hypotenuse=13\Rightarrow \text{hypotenuse} = 13
Therefore, sin⁡B=513\sin B = \dfrac{5}{{13}} and sec⁡B=1312\sec B = \dfrac{{13}}{{12}}

Put the value of the trigonometric function in the given equation. We get,
tan⁡2B−sin⁡2B=sin⁡4Bsec⁡2B{\tan ^2}B - {\sin ^2}B = {\sin ^4}B{\sec ^2}B
⇒(512)2+(513)2=(513)4(1312)2\Rightarrow {\left( {\dfrac{5}{{12}}} \right)^2} + {\left( {\dfrac{5}{{13}}} \right)^2} = {\left( {\dfrac{5}{{13}}} \right)^4}{\left( {\dfrac{{13}}{{12}}} \right)^2}
25144−25169=62524336\dfrac{{25}}{{144}} - \dfrac{{25}}{{169}} = \dfrac{{625}}{{24336}}
Solving the left side of the equation. We get,
62524336=62524336\dfrac{{625}}{{24336}} = \dfrac{{625}}{{24336}}
Hence, the left side of the equation is equal to the right side of the equation.
Hence proved.

Note: We can also use trigonometric identities such as 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta and sin⁡2θ−cos⁡2θ=1{\sin ^2}\theta - {\cos ^2}\theta = 1. To find the value of trigonometric function. As by tan⁡B\tan B we can find the value of sec⁡B\sec B and by using sec⁡B\sec B we can find the value of cos⁡B\cos B and by cos⁡B\cos B we can find the value of sin⁡B\sin B by using these trigonometric identities.