Question

Tan+sec-1/tan-sec+1=1+sin/cos

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

Answer:

Hey Mate,

We have to prove LHS = RHS

Numerator = tan theta + sec theta -1 
= sin theta/cos theta + 1/cos theta - 1 
= (sin theta + 1)/cos theta -1 

denominator = (sin theta -1)/cos theta +1 

multiply both by cos theta 

numerator = sin theta +1 - cos theta 
= 2 sin theta/2 cos theta/2 + 2 sin^ theta/2 

denominator = 2 sin theta/2 cos theta/2 - 2 sin ^2 theta/2 

divide 

number = (cos theta/2 + sin theta/2)/(cos theta/2 - sin theta/2) 

= (cos theta/2 + sin theta/2)^2/(cos^2 theta/2- sin ^2 theta/2) 
= (cos^2 theta/2+ sin^2 theta/2 + 2 cos theta/2 sin theta/2)/(cos theta) 
= (1+ sin theta)/ cos theta

= sec theta+ tan theta
= 1/cos theta + sin theta / cos theta
= 1+ sin theta/ cos theta


Hence Proved