Question

If $\cos x = \frac{1}{\sqrt{1 + t^{2}}}$ and $\sin y = \frac{t}{\sqrt{1 + t^{2}}}$, then $\frac{dy}{dx}$=

• A. – 1
• B. $\frac{1 - t}{1 + t^{2}}$
• C. $\frac{1}{1 + t^{2}}$
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Obviously $x = \cos^{- 1}\frac{1}{\sqrt{1 + t^{2}}}$ and $y = \sin^{- 1}\frac{t}{\sqrt{1 + t^{2}}}$

⇒ $x = \tan^{- 1}t$ and $y = \tan^{- 1}t$ ⇒ $y = x$ ⇒ $\frac{dy}{dx} = 1$.