Question

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

Answer 1

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.

Complete step-by-step answer:

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.