Solve the following:
$3\left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right) = 5$
Answer
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Hint – In this question use the trigonometric identity that $1 + {\tan ^2}\theta = {\sec ^2}\theta $. Convert ${\sec ^2}\theta $ in the inner bracket part to make the terms all in ${\tan ^2}\theta $, solve for $\tan \theta $ and thus the value of $\theta$ can be obtained. This will help to get the right answer.
Complete step-by-step answer:
Given trigonometric equation
$3\left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right) = 5$
As we know that ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$ \Rightarrow {\sec ^2}\theta = 1 + {\tan ^2}\theta $
Now substitute this value in given equation we have,
$ \Rightarrow 3\left( {1 + {{\tan }^2}\theta + {{\tan }^2}\theta } \right) = 5$
$ \Rightarrow 3 + 6{\tan ^2}\theta = 5$
$ \Rightarrow 6{\tan ^2}\theta = 5 - 3 = 2$
$ \Rightarrow {\tan ^2}\theta = \dfrac{2}{6} = \dfrac{1}{3}$
Now take square root on both sides we have,
$ \Rightarrow \tan \theta = \sqrt {\dfrac{1}{3}} = \dfrac{1}{{\sqrt 3 }}$
Now as we know that $\tan {30^o} = \dfrac{1}{{\sqrt 3 }}$
$ \Rightarrow \tan \theta = \tan {30^o}$
$ \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 ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\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 ${\tan ^2}\theta $ we could have changed into ${\sec ^2}\theta $ as well. The process would have been the same and thus the answer obtained will be the same too.
Complete step-by-step answer:
Given trigonometric equation
$3\left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right) = 5$
As we know that ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$ \Rightarrow {\sec ^2}\theta = 1 + {\tan ^2}\theta $
Now substitute this value in given equation we have,
$ \Rightarrow 3\left( {1 + {{\tan }^2}\theta + {{\tan }^2}\theta } \right) = 5$
$ \Rightarrow 3 + 6{\tan ^2}\theta = 5$
$ \Rightarrow 6{\tan ^2}\theta = 5 - 3 = 2$
$ \Rightarrow {\tan ^2}\theta = \dfrac{2}{6} = \dfrac{1}{3}$
Now take square root on both sides we have,
$ \Rightarrow \tan \theta = \sqrt {\dfrac{1}{3}} = \dfrac{1}{{\sqrt 3 }}$
Now as we know that $\tan {30^o} = \dfrac{1}{{\sqrt 3 }}$
$ \Rightarrow \tan \theta = \tan {30^o}$
$ \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 ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\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 ${\tan ^2}\theta $ we could have changed into ${\sec ^2}\theta $ as well. The process would have been the same and thus the answer obtained will be the same too.
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