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Question: Solve the following: \(3\left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right) = 5\)...

Solve the following:
3(sec2θ+tan2θ)=53\left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right) = 5

Explanation

Solution

Hint – In this question use the trigonometric identity that 1+tan2θ=sec2θ1 + {\tan ^2}\theta = {\sec ^2}\theta . Convert sec2θ{\sec ^2}\theta in the inner bracket part to make the terms all in tan2θ{\tan ^2}\theta , solve for tanθ\tan \theta and thus the value of θ\theta can be obtained. This will help to get the right answer.

Complete step-by-step answer:
Given trigonometric equation
3(sec2θ+tan2θ)=53\left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right) = 5
As we know that sec2θtan2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1
sec2θ=1+tan2θ\Rightarrow {\sec ^2}\theta = 1 + {\tan ^2}\theta
Now substitute this value in given equation we have,
3(1+tan2θ+tan2θ)=5\Rightarrow 3\left( {1 + {{\tan }^2}\theta + {{\tan }^2}\theta } \right) = 5
3+6tan2θ=5\Rightarrow 3 + 6{\tan ^2}\theta = 5
6tan2θ=53=2\Rightarrow 6{\tan ^2}\theta = 5 - 3 = 2
tan2θ=26=13\Rightarrow {\tan ^2}\theta = \dfrac{2}{6} = \dfrac{1}{3}
Now take square root on both sides we have,
tanθ=13=13\Rightarrow \tan \theta = \sqrt {\dfrac{1}{3}} = \dfrac{1}{{\sqrt 3 }}
Now as we know that tan30o=13\tan {30^o} = \dfrac{1}{{\sqrt 3 }}
tanθ=tan30o\Rightarrow \tan \theta = \tan {30^o}
θ=30o\Rightarrow \theta = {30^o}
So this is the required solution of the given expression.

Note – It is advised to remember the direct trigonometric identities like sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, 1 + cot2θ=cosec2θ{\text{1 + co}}{{\text{t}}^2}\theta = \cos e{c^2}\theta , as it helps saving a lot of time while solving problems of these kind. There can be another way of solving this problem, instead of changing the terms in tan2θ{\tan ^2}\theta we could have changed into sec2θ{\sec ^2}\theta as well. The process would have been the same and thus the answer obtained will be the same too.