Question

p = [1 + cosec (q) cos (q)] differentiate​

Answer 1

$\textbf{Given:}$

$P=1+cosec\,q\,cos\,q$

$\textbf{To find:}$

$\text{Derivative of P}$

$\textbf{Solution:}$

$\text{Consider,}$

$P=1+cosec\,q\,cos\,q$

$P=1+\dfrac{1}{sin\,q}\,cos\,q$

$P=1+\dfrac{cos\,q}{sin\,q}$

$P=1+cot\,q$

$\text{Clearly, P is a function of q}$

$\text{Differentiate with respect to q}$

$\dfrac{dP}{dq}=0+(-cosec^2q)$

$\dfrac{dP}{dq}=-cosec^2q$

$\textbf{Answer:}$

$\textbf{Derivative of \bf\,1+cosec\,q\,cos\,q is \bf-cosec^2q }$

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