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

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

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Grade:10

2 Answers

Arun
25763 Points
4 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
15 Points
2 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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