Prove that-tana+sina/tana-sina = seca+1/seca-1

Prove that-tanA+sinA/tanA-sinA = secA+1/secA-1

(tanA + sinA)/(tanA - sinA)

=(sinA/cosA + sin A)/(sinA/cosA - sinA)

=((sinA + sinAcosA)/cosA)/((sinA - sinAcosA)/cosA)

=(sinA(1+cosA))/(sinA(1-cosA))

=(1+cosA)/(1-cosA)

=(1+(1/secA))/(1-(1/secA))

=((secA + 1)/secA)/((secA-1)/secA)

=(secA+1)/(secA-1)

Step-by-step explanation:

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OK that is ur solution

Step-by-step explanation:

I first step the fill the value of tanA=sinA/cosA

in second step I take LCM

in third I cancel out the cosA

as cosA =1/secA

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answer:

step-by-step explanation:

simply use distance formula

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