1/1-sinA + 1/1+sinA = 2sec 2 A
Answers
Answered by
46
lhs= 1/(1- sinA) +1/(1+sinA)
=(1+sinA+1-sinA)/(1-sinA)(1+sinA)
=2/(1-sin^2A)here we used (a+b)(a-b)=a^2-b^2
=2/cos^2 A here 1-sin^2A=cos^2A
=2sec^2 A here 1/cos^2A= sec^2A
=rhs
=(1+sinA+1-sinA)/(1-sinA)(1+sinA)
=2/(1-sin^2A)here we used (a+b)(a-b)=a^2-b^2
=2/cos^2 A here 1-sin^2A=cos^2A
=2sec^2 A here 1/cos^2A= sec^2A
=rhs
Similar questions