JAMIA 2024 — Mathematics PYQ

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Question

JAMIA 2024 — Mathematics PYQ

JAMIA | Mathematics | 2024

$\sin(\tan^{-1} x), \ \text{where } |x| < 1, \text{ is equal to}$

Choose the correct answer:

Explanation

Solution:

Hume $\sin(\tan^{-1} x)$ ki value nikalni hai.

Step 1: Maan lijiye (Substitution)

Maan lijiye ki:

\theta = \tan^{-1} x

Iska matlab hai:

\tan \theta = x

Step 2: Right-angled triangle ka use karein

Hum jaante hain ki $\tan \theta = \frac{\text{Perpendicular (Lumb)}}{\text{Base (Aadhar)}}$ .

Yahan $\tan \theta = \frac{x}{1}$ hai, toh:

$\text{Perpendicular} = x$
$\text{Base} = 1$

Pythagoras theorem se Hypotenuse (Karn) nikalte hain:

\text{Hypotenuse} = \sqrt{(\text{Base})^2 + (\text{Perpendicular})^2}

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

Step 3: $\sin \theta$ ki value nikaalein

$\sin \theta$ ka formula hota hai $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ :

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

Step 4: $\theta$ ki value wapas rakhein

Kyonki humne $\theta = \tan^{-1} x$ maana tha, isliye:

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

Final Answer:

$\sin(\tan^{-1} x)$ barabar hai:

\mathbf{\frac{x}{\sqrt{1+x^{2}}}}