Question

1. tanA+cotA=secAcosecA [prove that]

Answer 1

Given,

TO PROVE :

$\tan(a) + \cot(a) = \sec(a) \csc(a)$

LHS :

$\tan(a) + \cot(a) = \dfrac{ \sin(a) }{ \cos(a) } + \dfrac{ \cos(a) }{ \sin(a) }$

$\implies \: \dfrac{ { \sin(a)}^{2} + { \cos(a) }^{2} }{ \sin(a) \cos(a) }$

we know that,

${ \sin(a) }^{2} + { \cos(a) }^{2} = 1$

$\implies \: \dfrac{1}{ \sin(a) \cos(a) }$

we know that,

$\dfrac{1}{ \sin(a) } = \csc(a)$

$\dfrac{1}{ \cos(a) } = \sec(a)$

$\implies \: \dfrac{1}{ \sin(a) \cos(a) } = \sec(a) \times \csc(a)$

LHS = RHS

Hence proved ,

$\tan(a) + \cot(a) = \sec(a) \csc(a)$

Answer 2

Answer:

LHS=sinA/cosA+cosA/sinA

LHS=sin^2A+cos^2A/cosA*sinA

LHS=1/cosA*sinA

LHS=1/cosA*1/sinA

LHS=secA*cosecA=RHS