Question

How do you verify ${\left( {\sin A - \cos A} \right)^2} = 1 - 2{\sin ^2}A\cot A$?

Answer 1

Hint: This question is related to the trigonometry, and we have to prove that the left hand side is equal to the right hand side of the expression, and this question can be solved by using algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and trigonometric identities i.e.,$\cot x = \dfrac{{\cos x}}{{\sin x}}$ and${\cos ^2}A + {\sin ^2}A = 1$, and by further simplifying we will get the result which is on the right hand.

Complete step by step answer:
Given function is ${\left( {\sin A - \cos A} \right)^2} = 1 - 2{\sin ^2}A\cot A$,
Now we have to prove that left hand side is equal to the right hand side of the equation, now take the expression on the left hand side i.e.,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2}$,
Now using the algebraic identity, ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, here $a = \sin A$ and $b = \cos A$, and by substituting the values in the identity we get,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2} = {\sin ^2}A + {\cos ^2}A - 2\sin A\cos A$,
Now using the trigonometric identities i.e., ${\cos ^2}A + {\sin ^2}A = 1$ , we get,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2} = 1 - 2\sin A\cos A$,
Now multiplying and dividing $\sin A$ to the right hand side of the equation we get,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2} = 1 - 2\sin A\cos A \times \dfrac{{\sin A}}{{\sin A}}$,
Now simplifying, we get,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2} = 1 - 2\sin A\left( {\sin A} \right) \times \dfrac{{\cos A}}{{\sin A}}$,
Now multiply like terms we get,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2} = 1 - 2{\sin ^2}A \times \dfrac{{\cos A}}{{\sin A}}$,
As we know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$, so write $\dfrac{{\cos A}}{{\sin A}}$ as $\cot A$, we get,
$ \Rightarrow {\left( {\sin A - \cos A} \right)^2} = 1 - 2{\sin ^2}A\cot A$,
Which is equal to the right hand side of the equation,
Hence proved.

Note: An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles. They are useful when solving questions with trigonometric functions and expressions. There are many trigonometric identities, here are some useful identities:
$\tan x = \dfrac{{\sin x}}{{\cos x}}$,
$\sec x = \dfrac{1}{{\cos x}}$,
$\cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}$ ,
${\sin ^2}x = 1 - {\cos ^2}x$,
${\cos ^2}x + {\sin ^2}x = 1$,
${\sec ^2}x - {\tan ^2}x = 1$,
${\csc ^2}x = 1 + {\cot ^2}x$.
${\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$,
${\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1$,
$\sin 2x = 2\sin x\cos x$,
$2{\cos ^2}x = 1 + \cos 2x$,
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$.