Question

Prove the following statement: $\left( \sin A+\cos A \right)\left( \cot A+\tan A \right)=\sec A+\csc A$

Answer 1

Hint: First we will write cotA and tanA in the form of sinA and cosA and then we will open the brackets and then we will the two expression, then we will rearrange some terms and after that we will use the trigonometric formulas and try to write the expression in simpler form to prove the above statement and just to be sure that the answer we have calculated is correct we will try to put some values in place of A to check if it’s matches or not.

Complete step-by-step answer:

The formulas that we are going to use to solve this question are:

$\begin{align}

  & \cot A=\dfrac{\cos A}{\sin A} \\

 & \tan A=\dfrac{\sin A}{\cos A} \\

 & \sec A=\dfrac{1}{\cos A} \\

 & \csc A=\dfrac{1}{\sin A} \\

\end{align}$

Now putting these formulas in $\left( \sin A+\cos A \right)\left( \cot A+\tan A \right)$ we get,

$\begin{align}

  & \left( \sin A+\cos A \right)\left( \dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A} \right) \\

 & =\left( \sin A+\cos A \right)\left( \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A\cos A} \right)

\\

 & =\left( \dfrac{\sin A+\cos A}{\sin A\cos A} \right) \\

 & =\left( \dfrac{1}{\cos A}+\dfrac{1}{\sin A} \right) \\

 & =\left( \sec A+\csc A \right) \\

\end{align}$

Hence by using all the formulas that we have stated above and then by rearranging some terms we have proved that,

$\left( \sin A+\cos A \right)\left( \cot A+\tan A \right)=\sec A+\csc A$

Note: It’s always better that we check if the answer that we have got by using the above formula we is correct or not to avoid some calculation mistake and for that we need to put some values in place of A to check whether it satisfies the above expression or not, so let’s check by putting A = $\dfrac{\pi }{4}$ ,

We get LHS = $\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right)\left( 1+1 \right)=2\sqrt{2}$ , now we will write RHS = $\left( \sqrt{2}+\sqrt{2} \right)=2\sqrt{2}$ ,

Hence from this we can easily say that the answer that we have calculated is correct and there is no calculation mistake. There are many ways to solve this question as one can start from RHS and then show that RHS = LHS.