prove that sin A(1+tanA)+cosA(1+cotA)=secA +cosecA
Answers
LHS= sinA(1+tanA)+cosA(1+cotA)
And RHS= secA+cosecA
Now simplifying LHS,
sinA(1+tanA)+cosA(1+cotA)
= sinA(1+sinA/cosA) + cosA(1+cosA/sinA)
{TanA=sinA/cosA and CotA=cosA/SinA}
= sinA(cosA+SinA/cosA) + cosA(sinA+cosA/sinA)
{By taking LCM}
=(sinAcosA+sin²A/cosA) + (sinAcosA+cos²A/sinA)
=sinA(sinAcosA+sin²A)+cosA(sinAcosA+cos²A)/sinAcosA
{By taking LCM}
sin²AcosA+sin³A+sinAcos²A+cos³A/sinAcosA
=(1-cos²A)(cosA)+sin³A+(sinA)(1-sin²A)+cos³A/sinAcosA
{Since sin²A=1-cos²A and cos²A=1-sin²A}
= cosA-cos³A+sin³A+sinA-sin³A+cos³A/sinAcosA
= cosA-cos³A+cos³A+sinA-sin³A+sin³A/sinAcosA
{By rearranging}
=sinA+cosA/sinAcosA
=1/cosA+1/sinA
=secA+cosecA=RHS
{Since 1/cosA=secA and 1/sinA=cosecA}
Hence proved....!!
I suggest you that you can solve this question on rough paper so that you can understand it easily.
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