If sinα -cosα /sinα +cosα where θ and α are (α ≠π /4) acute angles, then what is {2} sin θ equal to ?

Correct Answer is : sinα−cosαsin α − cos αsinα−cosα

Correct Answer is : sinα−cosαsin α − cos αsinα−cosα

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UPSC CDS I 2025 Mathematics Paper

Question : 33 of 100

Marks: +1, -0

\frac{sin\alpha -cos\alpha }{sin\alpha +cos\alpha }

θ

α

(\alpha ≠\frac{\pi }{4})

\sqrt{2} sin θ equal to ?

Solution:

Concept:

Use the relation between tangent and sine:

\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}

. Simplify using algebraic identities.

Explanation:

Given

\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}

We know

\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}

Let

k = \sin\alpha - \cos\alpha

and

m = \sin\alpha + \cos\alpha

. Then

\tan\theta = \frac{k}{m}

Compute

1+\tan^2\theta = 1 + \frac{k^2}{m^2} = \frac{m^2+k^2}{m^2}

Now

m^2+k^2 = (\sin\alpha+\cos\alpha)^2 + (\sin\alpha-\cos\alpha)^2 = 2(\sin^2\alpha+\cos^2\alpha) = 2

1+\tan^2\theta = \frac{2}{m^2}

, and

\sqrt{1+\tan^2\theta} = \frac{\sqrt{2}}{|m|}

. Since

\alpha

is acute,

m>0

, so

\sqrt{1+\tan^2\theta} = \frac{\sqrt{2}}{m}

Thus

\sin\theta = \frac{k/m}{\sqrt{2}/m} = \frac{k}{\sqrt{2}} = \frac{\sin\alpha - \cos\alpha}{\sqrt{2}}

Multiply both sides by

\sqrt{2}

\sqrt{2}\sin\theta = \sin\alpha - \cos\alpha

Answer:

Option A:

\sin\alpha - \cos\alpha

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