Question

How do you simplify $\tan \theta \left( \cot \theta +\tan \theta \right)$ ?

Answer 1

To simplify the above trigonometric expression we are going to substitute $\cot \theta =\dfrac{1}{\tan \theta }$ in the given expression and then simplify the expression written in the bracket first i.e. $\left( \cot \theta +\tan \theta \right)$ and then we are going to multiply this result with $\tan \theta$.

Complete step by step answer:
The trigonometric expression given in the above problem which we have to simplify is as follows:
$\tan \theta \left( \cot \theta +\tan \theta \right)$
We know that there is trigonometric conversion which is equal to:
$\cot \theta =\dfrac{1}{\tan \theta }$
Using the above relation in $\tan \theta \left( \cot \theta +\tan \theta \right)$ we get,
$\Rightarrow \tan \theta \left( \dfrac{1}{\tan \theta }+\tan \theta \right)$
Taking $\tan \theta$ as L.C.M in the bracket shown above we get,
$\Rightarrow \tan \theta \left( \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta } \right)$
Now, as you can see that $\tan \theta$ is written in the numerator and denominator of the above expression so it will be cancelled out from the numerator and denominator and we get,
$1+{{\tan }^{2}}\theta$
There is also a trigonometric identity in $\tan \theta \And \sec \theta$ which is equal to:
$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$
Using the above identity in $1+{{\tan }^{2}}\theta$ we get,
${{\sec }^{2}}\theta$

Hence, we have simplified the given trigonometric expression to ${{\sec }^{2}}\theta$.

Note: The alternate approach to simplify the above trigonometric expression is to write the following in the given expression:
$\begin{aligned} & \tan \theta =\dfrac{\sin \theta }{\cos \theta }; \\\ & \cot \theta =\dfrac{\cos \theta }{\sin \theta } \\\ \end{aligned}$
$\dfrac{\sin \theta }{\cos \theta }\left( \dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta } \right)$
Now, taking $\sin \theta \cos \theta$ as L.C.M in the denominator in the bracket of the above expression we get,
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }\left( \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\sin \theta \cos \theta } \right)$
As you can see that $\sin \theta$ is common in the numerator and the denominator in the above expression so we are going to cancel out $\sin \theta$ from the numerator and denominator and we get,
$\Rightarrow \dfrac{1}{\cos \theta }\left( \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\cos \theta } \right)$
We know that there is a trigonometric identity that:
${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Using the above relation in $\dfrac{1}{\cos \theta }\left( \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\cos \theta } \right)$ we get,
$\Rightarrow \dfrac{1}{\cos \theta }\left( \dfrac{1}{\cos \theta } \right)$
Multiplying $\cos \theta$ with $\cos \theta$ in the denominator of the above expression we get,
$\Rightarrow \dfrac{1}{{{\cos }^{2}}\theta }$
Rewriting the above trigonometric expression by taking as whole square of $\dfrac{1}{\cos \theta }$ we get,
$\Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}$
There is a relation that secant of any angle is equal to the reciprocal of cosine of an angle.
$\sec \theta =\dfrac{1}{\cos \theta }$
Using the above relation in ${{\left( \dfrac{1}{\cos \theta } \right)}^{2}}$ we get,
$\begin{aligned} & \Rightarrow {{\left( \sec \theta \right)}^{2}} \\\ & ={{\sec }^{2}}\theta \\\ \end{aligned}$