Correct Answer - Option 2 : -acosec
2x – (c × cosec x cot x)
Concept:
\(\rm \frac{\mathrm{d} (cotx)}{\mathrm{d} x} = -cosec^{2}x \)
\(\rm \frac{\mathrm{d} (cosecx)}{\mathrm{d} x} = - cotx\; cosecx\)
Calculation:
Given: f(x) = \(\rm \frac{(a\cos x + b \sin x + c)}{\sin x}\) = acot x + b + c cosec x
Now, f'(x) = \(\rm \frac{\mathrm{d} (a\cot x + b + c(cosec\; x))}{\mathrm{d} x}\)
⇒ -acosec2x – (c × cosec x cot x)
∴ The required value is -acosec2x – (c × cosec x cot x)