Question

If\[\cos A + \sin A = \sqrt 2 \cos A\] then $\cos A - \sin A = $
$
  \left( A \right)\,\,\sqrt 2 \sin A \\
  \left( B \right)\,\,\,2\sin A \\
  \left( C \right)\,\,\, - \sqrt 2 \sin A \\
  \left( D \right)\,\,\,\, - 2\sin A \\
 $

Answer 1

Hint: We need to remember the concept of trigonometric transformation formulae and trigonometric ratios and where the signs of cosine and sine will be positive and negative. To prove this we take the if condition part and then we solve the required part .

Complete Step by Step Solution:
The objective of the problem is to find the value of $\cos A - \sin A$
For this we considering if part of the given problem
Given that \[\cos A + \sin A = \sqrt 2 \sin A\]
From this we are finding sinA
That is by subtracting cosA on both sides we get
\[
  \cos A + \sin A - \cos A = \sqrt 2 \,\cos A - \cos A \\
   \Rightarrow \sin A = \sqrt 2 \,\cos A - \cos A \\
 \]
Taking cosine function as common on right hand side we get
\[
   \Rightarrow \sin A = \cos A\left( {\sqrt 2 \, - 1} \right) \\
   \Rightarrow \dfrac{{\sin A}}{{\cos A}} = \left( {\sqrt 2 \, - 1} \right) \\
   \Rightarrow \tan A = \left( {\sqrt 2 \, - 1} \right) \\
 \]
To find the value of cosA we are finding cot A because it is easy to find cot from the tan. We know that cot function is reciprocal of tan function. We get
$
   \Rightarrow \dfrac{1}{{\tan A}} = \dfrac{1}{{\sqrt 2 - 1}} \\
   \Rightarrow \cot A = \dfrac{1}{{\sqrt 2 - 1}} \\
 $
Now rationalizing the above term . that is multiplying and dividing with $\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}}$
Now we get ,
$
   \Rightarrow \cot A = \dfrac{1}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}} \\
   \Rightarrow \cot A = \dfrac{{\sqrt 2 + 1}}{{{{\left( {\sqrt 2 } \right)}^2} - {1^2}}} \\
   \Rightarrow \cot A = \dfrac{{\sqrt 2 + 1}}{{2 - 1}} \\
   \Rightarrow \cot A = \dfrac{{\sqrt 2 + 1}}{1} \\
   \Rightarrow \cot A = \sqrt 2 + 1 \\
 $
Now cot A can be written as
$ \Rightarrow \dfrac{{\cos A}}{{\sin A}} = \sqrt 2 + 1$
From the above equation we can write cos A as
$
   \Rightarrow \cos A = \sin A\left( {\sqrt 2 + 1} \right) \\
   \Rightarrow \cos A = \sin A\left( {\sqrt 2 } \right) + \sin A \\
 $
Now in the above equation we are subtracting sinA on both sides we get
$
  \cos A - \sin A = \sin A\left( {\sqrt 2 } \right) + \sin A - \sin A \\
   \Rightarrow \cos A - \sin A = \sqrt 2 \sin A \\
 $
Therefore, $\cos A - \sin A = \sqrt 2 \sin A$

Thus option A) is the correct.

Note:
The sine trigonometric function is positive in first and second quadrants in the coordinate axis and in the remaining it is negative only. And for the tan trigonometric function it is positive in the first and third quadrants of the coordinate system. For this type of problem we should remember the trigonometric ratios and formulae. And one should be careful about the algebraic signs and trigonometric signs.