CUET UG 2025 — Mathematics Calculus

Let $\theta = \tan^{-1}x$θ=tan−1x

By definition, $\tan \theta = x$tanθ=x

This can be written as:

$$\tan \theta = \frac{x}{1}$$tanθ=1x​

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Construct a right triangle where:

Opposite side $= x$=x

Adjacent side $= 1$=1

This gives $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1} = x$tanθ=adjacentopposite​=1x​=x

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Using the Pythagorean theorem:

$$\text{Hypotenuse} = \sqrt{x^2 + 1^2}$$Hypotenuse=x2+12​

$$\text{Hypotenuse} = \sqrt{x^2 + 1}$$Hypotenuse=x2+1​

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For the sine ratio:

$$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$$sinθ=hypotenuseopposite​

$$\sin \theta = \frac{x}{\sqrt{1 + x^2}}$$sinθ=1+x2​x​

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Since $\theta = \tan^{-1}x$θ=tan−1x:

$$\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$$sin(tan−1x)=1+x2​x​