Question

Prove the following the statement: $\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\dfrac{\sec A}{{{\csc }^{2}}A}-\dfrac{\csc A}{{{\sec }^{2}}A}$

Answer 1

Hint: First we are going to write sec and cosec in terms of sin and cos and then we are going to prove the above statement by using some trigonometric formulas and then we will start from RHS and then by using the formulas we will show that it is equal to LHS.

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}$


Let’s start by solving from RHS by putting these values.


$\begin{align}


  & \dfrac{\sec A}{{{\csc }^{2}}A}-\dfrac{\csc A}{{{\sec }^{2}}A} \\


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


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


 & =\dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A\cos A} \\


\end{align}$


Now we are going to use the formula:


${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$


We are also going to use:


${{\sin }^{2}}x+{{\cos }^{2}}x=1$


Using this we get,


$\begin{align}


  & =\dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A\cos A} \\


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


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


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


\end{align}$


Now we will write 1 as ${{\sin }^{2}}x+{{\cos }^{2}}x$ ,


$\begin{align}


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


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


 & =\left( \sin A-\cos A \right)\left( \tan A+\cot A+1 \right) \\


\end{align}$


Hence using all the formulas we have shown LHS = RHS.

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. There are many ways to solve this question as one can start from LHS and then using some of the trigonometric formula one can show that RHS = LHS.