If x = θ + θ/θ - θ,π/4

Correct Answer is : sin⁡θ+cos⁡θθ + θsinθ+cosθ

Correct Answer is : sin⁡θ+cos⁡θθ + θsinθ+cosθ

Test Index

UPSC CDS 2 2023 Math Paper

Question : 28 of 100

Marks: +1, -0

\tan x = \frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta},\frac{\pi}{4} < \theta < \frac{\pi}{2}

, then what is √2 sin x equal to?

Solution:

Concept:

We use the right triangle definition of tangent and sine, then simplify using the Pythagorean identity.

Explanation:

Given

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

and

\frac{\pi}{4} < \theta < \frac{\pi}{2}

, both numerator and denominator are positive.

Treat

\tan x

\frac{\text{opposite}}{\text{adjacent}}

in a right triangle.

Let opposite

= \sin\theta + \cos\theta

and adjacent

= \sin\theta - \cos\theta

Compute the hypotenuse

H

using Pythagoras:

H = \sqrt{(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2}

Expand:

H = \sqrt{\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta}

Cancel

2\sin\theta\cos\theta

H = \sqrt{2(\sin^2\theta + \cos^2\theta)}

Since

\sin^2\theta + \cos^2\theta = 1

H = \sqrt{2}

Now

\sin x = \frac{\text{opposite}}{H} = \frac{\sin\theta + \cos\theta}{\sqrt{2}}

Multiply by

\sqrt{2}

\sqrt{2}\sin x = \sqrt{2} \cdot \frac{\sin\theta + \cos\theta}{\sqrt{2}} = \sin\theta + \cos\theta

Answer:

Option A:

\sin\theta + \cos\theta

Go to Question:

Prev Question

Next Question