Question

Prove the following statement: $\left( 1+\cot A-\csc A \right)\left( 1+\tan A+\sec A \right)=2$

Answer 1

Hint: First we will open the brackets and multiply each of the given terms and then we are going to prove the above statement by using some trigonometric formulas and then we will start from LHS and then by using the formulas we will show that it is equal to RHS.

Complete step-by-step answer:

Let’s start by solving from LHS,

$\left( 1+\cot A-\csc A \right)\left( 1+\tan A+\sec A \right)$

$=1+\tan A+\sec A+\cot A+\cot A\tan A+\cot A\sec A-\csc A-\csc A\tan A-\csc A\sec A$
Now we are going to use some formula,

$\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}$

After substituting the values we get,

$\begin{align}

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

 & \cot A\sec A=\left( \dfrac{1}{\cos A} \right)\left( \dfrac{\cos A}{\sin A} \right)=\csc A \\

 & \csc A\tan A=\left( \dfrac{1}{\sin A} \right)\left( \dfrac{\sin A}{\cos A} \right)=secA \\

\end{align}$

Therefore, the value of LHS becomes,

 $\begin{align}

  & =1+\tan A+\sec A+\cot A+1+\csc A-\csc A-secA-\csc A\sec A \\

 & =2+\left( \dfrac{\sin A}{\cos A} \right)+\left( \dfrac{\cos A}{\sin A} \right)-\dfrac{1}{\sin A\cos A} \\

\end{align}$

Now we also know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ ,

$\begin{align}

  & =2+\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}-\dfrac{1}{\sin A\cos A} \\

 & =2+\dfrac{1}{\sin A\cos A}-\dfrac{1}{\sin A\cos A} \\

 & =2 \\

\end{align}$

And hence we have shown that LHS = RHS by using the above formula.

Hence Proved.

Note: It’s always better that we check if the answer that we have got by using the above formula 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. There are a bunch of trigonometric formulas that should be kept in mind while solving these questions and if we use some different set of formulas then that will be another method to solve this question, but at some point we can see that they are nearly equal.