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Trigonometry> Prove that tan a + sec a - 1/tan a - sec ...

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Prove that tan a + sec a - 1/tan a - sec a + 1 =1 + sin a / cos a

Sukanya , 7 Years ago

Grade 10

2 Answers

Arun

Last Activity: 7 Years ago

(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cos A

multiply LHS by cosA /cosA to get
(sinA+1-cosA) / (sinA-1+cosA)

multiply again by cosA/cosA to get
(sinA.cosA+cosA-cos^2A) / cosA(sinA-1+cosA)

= ( cosA(1+sinA) - (1-sin^2A) ) / cosA(sinA-1+cosA)
= ( cosA(1+sinA) - (1+sinA)(1-sinA) ) / cosA(sinA-1+cosA)
= ( (1+sinA)(cosA-1+sinA) ) / cosA(sinA-1+cosA)
= (1+sinA)/cosA

Ibrahim Gbemisola

Last Activity: 6 Years ago

(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cos A

Recall that: sec^2A-tan^2A = 1

LHS = tanA+secA-(sec^2A-tan^2A)/tanA-secA+1

tanA+secA-[(secA+tanA)(secA-tanA)]/tanA-secA+1

(tanA+secA)[1-(secA-tanA)]/tanA-secA+1 =

tanA+secA

tanA+secA = sinA/cosA+1/cosA = 1+sinA/cosA

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