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# If $\tan \theta \tan {40^0} = 1$, then find the value of $\theta$.  Verified
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Hint: In this question use the property of tan$\theta$ which is $\tan \theta = \cot \left( {{{90}^0} - \theta } \right)$, $\dfrac{1}{{\tan x}} = \cot x$, so use these properties of tan$\theta$ to reach the solution of the question.

$\tan \theta \tan {40^0} = 1$
$\Rightarrow \tan \theta = \dfrac{1}{{\tan {{40}^0}}}$
Now as we know that $\dfrac{1}{{\tan x}} = \cot x$, so use this property in above equation we have,
$\Rightarrow \tan \theta = \dfrac{1}{{\tan {{40}^0}}} = \cot {40^0}$
$\Rightarrow \tan \theta = \cot {40^0}$
Now as we know that $\tan \theta = \cot \left( {{{90}^0} - \theta } \right)$, so substitute this value in above equation we have,
$\Rightarrow \cot \left( {{{90}^0} - \theta } \right) = \cot {40^0}$
$\begin{gathered} {90^0} - \theta = {40^0} \\ \Rightarrow \theta = {90^0} - {40^0} = {50^0} \\ \end{gathered}$
So, this is the required value of $\theta$.
Note: In such types of questions the key concept we have to remember is that we always recall all the basic trigonometric properties which are stated above, then using these properties simplify the given equation, we will get the required value of $\theta$.