Question

If $\sin \theta + \cos \theta = \sqrt 2$, then find the value of $\tan \theta + \cot \theta$.

Answer 1

Here we will firstly square both sides of the given equation to get the value of $\sin \theta \cos \theta$. Then we will divide the given equation by $\cos \theta$ to get the equation in terms of $\tan \theta$. Then again we will divide the given equation by $\sin \theta$ to get the equation in terms of $\cot \theta$. We will then add these equations with $\tan \theta$ and $\cot \theta$ to get the value of $\tan \theta + \cot \theta$.

Complete step by step solution:
It is given that $\sin \theta + \cos \theta = \sqrt 2$………………$\left( 1 \right)$
Now, we will be squaring both side of the equation $\left( 1 \right)$. Therefore, we get
$\Rightarrow {\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\sqrt 2 } \right)^2}$
$\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 2$
We know from the trigonometric properties that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get
$\Rightarrow 1 + 2\sin \theta \cos \theta = 2$
Subtracting the lie terms, we get
$\begin{array}{l} \Rightarrow 2\sin \theta \cos \theta = 2 - 1\\\ \Rightarrow 2\sin \theta \cos \theta = 1\end{array}$
Dividing both sides by 2, we get
$\Rightarrow \sin \theta \cos \theta = \dfrac{1}{2}$ ………………$\left( 2 \right)$
Now we will divide the equation $\left( 1 \right)$ by $\cos \theta$. Therefore, we get
$\Rightarrow \dfrac{{\sin \theta + \cos \theta }}{{\cos \theta }} = \dfrac{{\sqrt 2 }}{{\cos \theta }}$
$\Rightarrow \tan \theta + 1 = \dfrac{{\sqrt 2 }}{{\cos \theta }}$………………$\left( 3 \right)$
Now we will divide the equation $\left( 1 \right)$ by $\sin \theta$. Therefore, we get
$\Rightarrow \dfrac{{\sin \theta + \cos \theta }}{{\sin \theta }} = \dfrac{{\sqrt 2 }}{{\sin \theta }}$
$\Rightarrow 1 + \cot \theta = \dfrac{{\sqrt 2 }}{{\sin \theta }}$………………$\left( 4 \right)$
Now adding the equation $\left( 3 \right)$ and equation $\left( 4 \right)$, we get
$\Rightarrow \tan \theta + 1 + 1 + \cot \theta = \dfrac{{\sqrt 2 }}{{\cos \theta }} + \dfrac{{\sqrt 2 }}{{\sin \theta }}$
Adding the like terms, we get
$\Rightarrow \tan \theta + \cot \theta + 2 = \dfrac{{\sqrt 2 }}{{\cos \theta }} + \dfrac{{\sqrt 2 }}{{\sin \theta }}$
Taking LCM on the right side of the equation, we get
$\Rightarrow \tan \theta + \cot \theta + 2 = \dfrac{{\sqrt 2 \left( {\sin \theta + \cos \theta } \right)}}{{\sin \theta \cos \theta }}$
Subtracting 2 from both the sides, we get
$\Rightarrow \tan \theta + \cot \theta = \dfrac{{\sqrt 2 \left( {\sin \theta + \cos \theta } \right)}}{{\sin \theta \cos \theta }} - 2$
Now by using the equation $\left( 1 \right)$ we will put the value of $\sin \theta + \cos \theta$ in the equation and also by using the equation $\left( 2 \right)$ we will put the value of $\sin \theta \cos \theta$ in the equation, we get
$\Rightarrow \tan \theta + \cot \theta = \dfrac{{\sqrt 2 \left( {\sqrt 2 } \right)}}{{\dfrac{1}{2}}} - 2$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow \tan \theta + \cot \theta = \dfrac{2}{{\dfrac{1}{2}}} - 2\\\ \Rightarrow \tan \theta + \cot \theta = 4 - 2\\\ \Rightarrow \tan \theta + \cot \theta = 2\end{array}$
Hence, $\tan \theta + \cot \theta$ is equal to 2.

Note: When we add two equations, then the terms on the left side of both the equations are added and terms on the right side of both the equations are added. Here, if we didn’t find the square of the given equation, then we will not be able to find the value of $\sin \theta \cos \theta$. Hence we would not be able to solve the question.
We should note that the ratio of the $\sin \theta$ and $\cos \theta$ is equal to the $\tan \theta$. Also the ratio of $\cos \theta$ and $\sin \theta$ is equal to $\cot \theta$.
That is $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$.