QUESTION

# Prove that $\left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right) = \dfrac{{\sec A}}{{\cos e{c^2}A}} - \dfrac{{\cos ecA}}{{{{\sec }^2}A}} = \sin A\tan A - \cot A\cos A$

Hint: In this question first of all divide the given equation into 3 parts. Then solve part 1 to prove part 1 and 2 are equal and then solve part 2 to prove that part 2 and 3 are equal by using simple trigonometric ratios. By this we can say that all the three parts are equal to each other which is our required solution.

Divide the equation into three 3 parts do solve it easily.
Given LHS i.e., part 1 is $\left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right)$
$\Rightarrow \left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right)$
Writing the terms of cot and tan in terms of sin and cos we have
$\Rightarrow \left( {1 + \dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\cos A}}} \right)\left( {\sin A - \cos A} \right) \\ \Rightarrow \left( {\dfrac{{\sin A\cos A + {{\cos }^2}A + {{\sin }^2}A}}{{\sin A\cos A}}} \right)\left( {\sin A - \cos A} \right) \\ \Rightarrow \left[ {\dfrac{{\left( {\sin A\cos A + {{\cos }^2}A + {{\sin }^2}A} \right)\left( {\sin A - \cos A} \right)}}{{\sin A\cos A}}} \right] \\$
Multiplying the terms inside the brackets, we have
$\Rightarrow \left[ {\dfrac{{{{\sin }^2}A\cos A + \sin A{{\cos }^2}A + {{\sin }^3}A - \sin A{{\cos }^2}A - {{\cos }^3}A - {{\sin }^2}A\cos A}}{{\sin A\cos A}}} \right]$
Cancelling the common terms, we have
$\Rightarrow \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\sin A\cos A}}$
Splitting the terms, we have
$\Rightarrow \dfrac{{{{\sin }^3}A}}{{\sin A\cos A}} - \dfrac{{{{\cos }^3}A}}{{\sin A\cos A}} \\ \Rightarrow \dfrac{{{{\sin }^2}A}}{{\cos A}} - \dfrac{{{{\cos }^2}A}}{{\sin A}} \\$
Which can be written as
$\Rightarrow \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}} \\ \therefore \left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right) = \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}}......................................\left( 1 \right) \\$
Now consider the part 2 i.e., $\dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}}$
$\Rightarrow \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}} \\ \Rightarrow \dfrac{{\sec A}}{{\operatorname{cosec} A\operatorname{cosec} A}} - \dfrac{{\operatorname{cosec} A}}{{\sec A\sec A}} \\ \Rightarrow \dfrac{1}{{\operatorname{cosec} A}} \times \dfrac{{\sec A}}{{\operatorname{cosec} A}} - \dfrac{{\operatorname{cosec} A}}{{\sec A}} \times \dfrac{1}{{\sec A}} \\ \Rightarrow \sin A\tan A - \cot A\cos A \\ \therefore \dfrac{{\sec A}}{{{\text{cose}}{{\text{c}}^2}A}} - \dfrac{{\operatorname{cosec} A}}{{{{\sec }^2}A}} = \sin A\tan A - \cot A\cos A....................................................\left( 2 \right) \\$
From equation (1) and (2) we have
$\left( {1 + \cot A + \tan A} \right)\left( {\sin A - \cos A} \right) = \dfrac{{\sec A}}{{\cos e{c^2}A}} - \dfrac{{\cos ecA}}{{{{\sec }^2}A}} = \sin A\tan A - \cot A\cos A$
Hence proved.

Note: Here we have used the trigonometric ratio conversions such as $\dfrac{1}{{\operatorname{cosec} A}} = \sin A$, $\dfrac{{\sec A}}{{\operatorname{cosec} A}} = \dfrac{{\sin A}}{{\cos A}} = \tan A$, $\dfrac{{\operatorname{cosec} A}}{{secA}} = \dfrac{{\cos A}}{{\sin A}} = \cot A$, $\dfrac{1}{{\sec A}} = \cos A$. While multiplying the terms inside the brackets make sure that you have written all the terms multiplied in it.