KCET · Maths · Trigonometric Ratios & Identities
If \(y=\sin x \cdot \sin 2 x \cdot \sin 3 x \ldots \sin n x\), then \(y^{\prime}\) is
- A \(\sum_{k=1}^{n} k \tan k x\)
- B \(y \cdot \sum_{k=1}^{n} k \cot k x\)
- C \(y \cdot \sum_{k=1}^{n} k \tan k n\)
- D \(\sum_{k=1}^{n} \cot k x\)
Answer & Solution
Correct Answer
(B) \(y \cdot \sum_{k=1}^{n} k \cot k x\)
Step-by-step Solution
Detailed explanation
Given, \(y=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \ldots . \sin n x\).
Taking log on both sides,
\(\log y=\log \sin x+\log \sin 2 x+\ldots+\log \sin n x\)
Differentiating w.r.t. \(x\),
\[
\frac{1}{y} \cdot \frac{d y}{d x}=1 \cdot \cot x+2 \cot 2 x+\ldots+n \cot n x
\]
\(\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=y \cdot \sum_{k=1}^{n} k \cot k x \\ \Rightarrow & y^{\prime}=y \cdot \sum_{k=1}^{n} k \cot k x\end{array}\)
Taking log on both sides,
\(\log y=\log \sin x+\log \sin 2 x+\ldots+\log \sin n x\)
Differentiating w.r.t. \(x\),
\[
\frac{1}{y} \cdot \frac{d y}{d x}=1 \cdot \cot x+2 \cot 2 x+\ldots+n \cot n x
\]
\(\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=y \cdot \sum_{k=1}^{n} k \cot k x \\ \Rightarrow & y^{\prime}=y \cdot \sum_{k=1}^{n} k \cot k x\end{array}\)
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