Question

derivative of (1+tanq)/(sinq)​

Answer 1

Explanation :

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$\star\: \rm \green{ \dfrac{d}{dx} \bigg ( \dfrac{1 + \tan(x) }{ \sin(x) } \bigg)}$

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Apply the quotient rule,

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$\bold{ \bigg(\dfrac{f}{g} \bigg)' = \dfrac{f'.g - g'.f}{g {}^{2} } }$

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$\longrightarrow \rm \: \dfrac{ \dfrac{d}{dx} \bigg( 1 + \tan(x) \bigg ) \sin(x) - \dfrac{d}{dx} \bigg( \sin(x) (1 + \tan(x) \bigg) }{ \bigg( \sin(x) \bigg) {}^{2} } \\ \\ \\ \longrightarrow \rm \: \dfrac{ \bigg( \sec {}^{2} (x) \bigg ) \sin(x) - \bigg( \cos(x) (1 + \tan(x) \bigg) }{ \bigg( \sin(x) \bigg) {}^{2} } \\ \\ \\ \longrightarrow \rm \dfrac{ \sec(x) \times \sec(x) \times \sin(x) - \cos(x) \bigg( 1 + \dfrac{ \sin(x) }{ \cos(x) } \bigg) }{sin {}^{2} (x)} \\ \\ \\ \longrightarrow \rm \dfrac{ \sec(x) \times \dfrac{sin(x)}{cos(x)} - \cos(x) \bigg(\dfrac{ \cos(x) + \sin(x) }{ \cos(x) } \bigg) }{sin {}^{2} (x)} \\ \\ \\ \longrightarrow \rm \red{ \dfrac{ \sec(x) \times \tan(x) - \cos(x) - \sin(x) }{sin {}^{2} (x)} }$