NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

If $\sin x + a \cos x = b$ , then what is the expression for $|a \sin x - \cos x|$ in terms of $a$ and $b$ ?

Choose the correct answer:

Explanation

Solution

1. Set up the equations:

Let the required expression be $y$ :

y = |a \sin x - \cos x|

The given equation is:

b = \sin x + a \cos x

2. Square both equations:

First equation squared:

b^2 = (\sin x + a \cos x)^2

b^2 = \sin^2 x + a^2 \cos^2 x + 2a \sin x \cos x \quad \dots \text{(i)}

Second equation squared:

y^2 = (a \sin x - \cos x)^2

y^2 = a^2 \sin^2 x + \cos^2 x - 2a \sin x \cos x \quad \dots \text{(ii)}

3. Add equations (i) and (ii):

b^2 + y^2 = (\sin^2 x + a^2 \cos^2 x + 2a \sin x \cos x) + (a^2 \sin^2 x + \cos^2 x - 2a \sin x \cos x)

b^2 + y^2 = \sin^2 x + \cos^2 x + a^2 \sin^2 x + a^2 \cos^2 x

b^2 + y^2 = (\sin^2 x + \cos^2 x) + a^2(\sin^2 x + \cos^2 x)

4. Use the identity $\sin^2 x + \cos^2 x = 1$ :

b^2 + y^2 = 1 + a^2(1)

b^2 + y^2 = 1 + a^2

5. Solve for $y$ :

y^2 = a^2 - b^2 + 1

y = \sqrt{a^2 - b^2 + 1}