Question

A tangent is drawn from the origin \((0,0)\) to the curve \(y=\sin x\). If \((x,y)\) is the point of contact and \(x\neq 0\), then which of the following relations is true?

• A. \(\frac{1}{y^{2}}-\frac{1}{x^{2}}=1\)
• B. \(\frac{1}{y^{2}}+\frac{1}{x^{2}}=1\)
• C. \(\frac{1}{x^{2}}-\frac{1}{y^{2}}=1\)
• D. \(\frac{1}{x^{2}}+\frac{1}{y^{2}}=-1\)

Hint:

Equate the slope of the tangent at the point of contact with the slope of the line joining it to the origin.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The tangent at \((x,y)\) has slope \(\cos x\). Since it passes through the origin, its slope is also \(\dfrac yx\).

Step 2: Equate the slopes
\(\cos x=\dfrac{\sin x}{x}\), so \(\tan x=x\) (with \(y=\sin x\)).

Step 3: Use an identity
\(\dfrac1{\sin^2x}=1+\cot^2x=1+\dfrac1{\tan^2x}\).

Step 4: Substitute
With \(y=\sin x\) and \(\tan x=x\): \(\dfrac1{y^2}=1+\dfrac1{x^2}\), so \(\dfrac1{y^2}-\dfrac1{x^2}=1\).

Final Answer:
The relation is \(\dfrac1{y^2}-\dfrac1{x^2}=1\). \[ \boxed{\text{(A) }\dfrac{1}{y^{2}}-\dfrac{1}{x^{2}}=1} \]