Question:
If \(y = cot^{-1}(\frac{1+sin5x}{cos5x})\), then the value of \(\frac{dy}{dx}\) is
If \(y = cot^{-1}(\frac{1+sin5x}{cos5x})\), then the value of \(\frac{dy}{dx}\) is
Show Hint
Rewrite (1 + sin t)/cos t as tan(pi/4 + t/2).
Updated On: Oct 1, 2026
- \(-5\)
- \(5\)
- \(-\frac{2}{5}\)
- \(-\frac{5}{2}\)
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The Correct Option is D
Solution and Explanation
Step 1: Simplify the Argument:
For any angle \(t\), \(\dfrac{1+\sin t}{\cos t}=\tan\left(\dfrac\pi4+\dfrac t2\right)\). With \(t=5x\):
\[ \frac{1+\sin5x}{\cos5x}=\tan\left(\frac\pi4+\frac{5x}2\right) \]
Step 2: Use the Inverse Identity:
\(\cot^{-1}(\tan\varphi)=\dfrac\pi2-\varphi\). So
\[ y=\frac\pi2-\frac\pi4-\frac{5x}2=\frac\pi4-\frac{5x}2 \]
Step 3: Differentiate:
\[ \frac{dy}{dx}=-\frac52 \]
Step 4: Check the Options:
(A) \(-5\) and (B) \(5\) forget the half-angle factor. (C) \(-2/5\) inverts the fraction. Only \(-5/2\) fits, option (D).
Final Answer:
The derivative is \(-\dfrac52\), option (D). \[ \boxed{\text{(D) } -\frac{5}{2}} \]
For any angle \(t\), \(\dfrac{1+\sin t}{\cos t}=\tan\left(\dfrac\pi4+\dfrac t2\right)\). With \(t=5x\):
\[ \frac{1+\sin5x}{\cos5x}=\tan\left(\frac\pi4+\frac{5x}2\right) \]
Step 2: Use the Inverse Identity:
\(\cot^{-1}(\tan\varphi)=\dfrac\pi2-\varphi\). So
\[ y=\frac\pi2-\frac\pi4-\frac{5x}2=\frac\pi4-\frac{5x}2 \]
Step 3: Differentiate:
\[ \frac{dy}{dx}=-\frac52 \]
Step 4: Check the Options:
(A) \(-5\) and (B) \(5\) forget the half-angle factor. (C) \(-2/5\) inverts the fraction. Only \(-5/2\) fits, option (D).
Final Answer:
The derivative is \(-\dfrac52\), option (D). \[ \boxed{\text{(D) } -\frac{5}{2}} \]
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