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Question

# Prove that the $\frac{d}{dx}\left(secx\right)=secx\mathrm{tan}x$

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Solution

## STEP 1 : Solving the LHS of the equationWe know that,$secx=\frac{1}{\mathrm{cos}x}$Taking the LHS of the equation and applying quotient rule to find the derivative of$⇒\frac{d}{dx}\left(secx\right)=\frac{d}{dx}\left(\frac{1}{\mathrm{cos}x}\right)$$⇒\frac{\frac{d}{dx}1.\left(\mathrm{cos}x\right)-\frac{d}{dx}\mathrm{cos}x.\left(1\right)}{{\left(\mathrm{cos}x\right)}^{2}}$Simplifying the above equation$⇒\frac{\left(0\right)\left(\mathrm{cos}x\right)-\left(-\mathrm{sin}x\right)}{{\left(\mathrm{cos}x\right)}^{2}}$$⇒\frac{\mathrm{sin}x}{\left(\mathrm{cos}x\right)\left(\mathrm{cos}x\right)}$$⇒\left(\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)\left(\frac{1}{\mathrm{cos}x}\right)$$⇒\mathrm{tan}xsecx$Thus, LHS = RHS.Hence proved.

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