Question

If \[ y=\frac{a\cos x+b\sin x+c}{\sin x}, \] then \[ \frac{dy}{dx}= \]

• A. \(-a\cosec^2x-c\cosec x\cot x\)
• B. \(-a\)
• C. \(-a\cosec^2x+b\sec^2x+c\cosec x\cot x\)
• D. \(a\cosec^2x-c\cosec x\cot x\)

Hint:

Before differentiating, simplify trigonometric fractions into \(\cot x\), \(\tan x\), \(\sec x\), or \(\cosec x\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a function \( y \) defined as a quotient of a sum of trigonometric terms.
The goal is to find its first derivative with respect to \( x \).
Instead of using the Quotient Rule directly, it is easier to split the fraction first and simplify the expression into basic trigonometric functions.
Step 2: Key Formula or Approach:
1. Simplify: \( \frac{U+V+W}{Z} = \frac{U}{Z} + \frac{V}{Z} + \frac{W}{Z} \).
2. Basic derivatives:
\( \frac{d}{dx}(\cot x) = -cosec^2x \).
\( \frac{d}{dx}(constant) = 0 \).
\( \frac{d}{dx}(cosecx) = -cosecx \cot x \).
Step 3: Detailed Explanation:

Rewrite the original function by dividing each term in the numerator by \( \sin x \):
\[ y = \frac{a \cos x}{\sin x} + \frac{b \sin x}{\sin x} + \frac{C}{\sin x} \]

Use trigonometric identities \( \cot x = \frac{\cos x}{\sin x} \) and \( \text{cosec } x = \frac{1}{\sin x} \):
\[ y = a \cot x + b + C \text{cosec } x \]

Now, differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(a \cot x) + \frac{d}{dx}(b) + \frac{d}{dx}(C \text{cosec } x) \]

Apply the power rule and basic trigonometric derivative rules:
The derivative of \( \cot x \) is \( -cosec^2x \).
The derivative of the constant \( b \) is 0.
The derivative of \( \text{cosec } x \) is \( -\text{cosec } x \cot x \).

Substituting these back into the expression:
\[ \frac{dy}{dx} = a(-cosec^2x) + 0 + C(-\text{cosec } x \cot x) \]
\[ \frac{dy}{dx} = -a \cdot cosec^2x - C \cdot \text{cosec } x \cot x \]

Step 4: Final Answer:
The derivative is \( -acosec^2x - C \text{cosec } x \cot x \).