Question

1/1-sinA + 1/1+sinA = 2sec2A

Answer 1

1/1-sinA+1/1+sinA
=(1+sinA+1-sinA)/{(1-sinA)(1+sinA)}
=2/(1-sin²A)
=2/cos²A [∵, sin²A+cos²A=1]
=2sec²A (Proved)

Answer 2

Answer:

Step-by-step explanation:

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