KCET · Maths · Differentiation
If \(\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a\), then \(\frac{d y}{d x}\) is
- A \(\frac{y-1}{x+1}\)
- B \(\frac{y+1}{x-1}\)
- C \(\frac{x-1}{y-1}\)
- D \(\frac{x-1}{y+1}\)
Answer & Solution
Correct Answer
(A) \(\frac{y-1}{x+1}\)
Step-by-step Solution
Detailed explanation
Given, \(\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a\)
\(\Rightarrow \quad \frac{1+x}{1-y}=\sec a\)
\(\Rightarrow \quad 1+x=(1-y) \sec a\)
\(\Rightarrow \quad y \sec a=\sec a-1-x\)
\(\Rightarrow \quad \frac{d y}{d x} \sec a=-1\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-1}{\sec a}=\frac{-1}{\left(\frac{1+x}{1-y}\right)}\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-(1-y)}{(1+x)} \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}\)
\(\Rightarrow \quad \frac{1+x}{1-y}=\sec a\)
\(\Rightarrow \quad 1+x=(1-y) \sec a\)
\(\Rightarrow \quad y \sec a=\sec a-1-x\)
\(\Rightarrow \quad \frac{d y}{d x} \sec a=-1\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-1}{\sec a}=\frac{-1}{\left(\frac{1+x}{1-y}\right)}\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-(1-y)}{(1+x)} \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}\)
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