Prove (Cos/1+sin)+(1+sin/cos)=2/Cos?

Prove cos/(1 + sin) + (1 + sin)/cos = 2/cos.

LHS
= cos/(1 + sin) + (1 + sin)/cos
= [cos^2 + (1 + sin)^2]/[cos(1 + sin)]
= [cos^2 + 1 + 2sin + sin^2]/[cos(1 + sin)]
= [sin^2 + cos^2 + 1 + 2sin]/[cos(1 + sin)]
= 2[1 + sin]/[cos(1 + sin)]
= 2/cos
= RHS

Thus, the equation is true.

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Tan u + cos u /(1+sin u) = tan u + (cos u /(1+sin u)) * ((1-sin u)/(1-sin u)) = tan u + cos u *(1-sin u)/(1-sin^2 u) = tan u + cos u * (1-sin u)/cos^2 u = tan u + (1-sin u)/cos u = tan u + (1/cos u) - (sin u/cos u) = tan u + sec u - tan u = sec u.

cosx/(1+sinx) + (1 + sinx)/(cosx)

=(cosx(cosx) + (1+sinx)^2)/((1+sinx)*cosx) (common denominator)

=cos^2x+1+2sinx+sin^2x)/((1+sinx)*cosx) (multiplied out terms)

=(1 + 1 + 2 sinx)/((1+sinx)(cosx)) (sin^2x + cos^2x = 1)

=2(1+sinx)/((1+sinx)(cosx)) (simplified)

=2/cosx

cosx/(1 + sinx) + (1 + sinx)/cosx
cos^2x/cosx(1 + sinx) + (1 + sinx)^2/cosx(1 + sinx)
(cos^2x + (1 + sinx)^2)/cosx(1 + sinx)
(cos^2x + 1 + 2sinx + sin^2x)/cosx(1 + sinx)
((cos^2x + sin^2x) + 1 + 2sinx)/cosx(1 + sinx)

recall cos^2x + sin^2x = 1

(1 + 1 + 2sinx)/cosx(1 + sinx)
2(1 + sinx)/cosx(1 + sinx)
2/cosx


Find a common denominator, and it becomes pretty clear from there when you notice both sin^2x and cos^2x are in now in the numerator.

you need to restate that it really does not make any sense