Question

Prove that $\left( \sin \alpha +\cos \alpha \right)\left( \tan \alpha +\cot \alpha \right)=\sec \alpha +\csc \alpha $.

Answer 1

Hint: We have a sum of two terms on the left-hand side of $\left( \tan \alpha +\cot \alpha \right)$. We simplify $\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }$ and $\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }$ on both numerator and denominator. Then we add them as the LCM of the denominators are the multiplication of them. Then we use the identity of $\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)=1$ to find the solution of the numerator. We also need to mention the conditions.

Complete step-by-step solution:
We simplify the addition part of $\left( \tan \alpha +\cot \alpha \right)$ by breaking it as $\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }$ and $\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }$. We replace the values as and get
$\begin{align}
  & \left( \tan \alpha +\cot \alpha \right) \\
 & =\dfrac{\sin \alpha }{\cos \alpha }+\dfrac{\cos \alpha }{\sin \alpha } \\
 & =\dfrac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{\sin \alpha \cos \alpha } \\
\end{align}$
We replace the value with $\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)=1$.
Therefore, $\left( \tan \alpha +\cot \alpha \right)=\dfrac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{\sin \alpha \cos \alpha }=\dfrac{1}{\sin \alpha \cos \alpha }$
We multiply $\dfrac{1}{\sin \alpha \cos \alpha }$ with $\left( \sin \alpha +\cos \alpha \right)$.
So, $\left( \sin \alpha +\cos \alpha \right)\times \dfrac{1}{\sin \alpha \cos \alpha }=\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }$.
We break the individual terms and get $\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=\dfrac{\sin \alpha }{\sin \alpha \cos \alpha }+\dfrac{\cos \alpha }{\sin \alpha \cos \alpha }$.
Bu eliminating common terms we get $\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=\dfrac{1}{\cos \alpha }+\dfrac{1}{\sin \alpha }$
We know that the relations for inverse are $\dfrac{1}{\cos \alpha }=\sec \alpha ,\dfrac{1}{\sin \alpha }=\csc \alpha $
The equation becomes $\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=\dfrac{1}{\cos \alpha }+\dfrac{1}{\sin \alpha }=\sec \alpha +\csc \alpha $.
Thus proved $\left( \sin \alpha +\cos \alpha \right)\left( \tan \alpha +\cot \alpha \right)=\sec \alpha +\csc \alpha $.

Note: It is important to remember the condition to eliminate the \[\cos \alpha ,\sin \alpha \] from both denominator and numerator. No domain is given for the variable $x$. The value of $\tan x\ne 0$ is essential. The simplified condition will be $x\ne n\pi ,n\in \mathbb{Z}$.