Question

how do you solve this problem?

1+sinØ/cosØ +cosØ/1+sinØ = 2secØ
and

cosß- cosß/1-tanß = sinßcosß/sinß-cosß

Answer 1

Since you did not use parentheses for your numerators and denominators, I spent most of my time figuring out what you probably mean

(1 + sin x)/( cos x) + cos x/(1+sin x) = 2/cos x

common denominator on left of
(1+sin x) cos x

[(1 +sin x)^2 +cos^2 x ]/common denom

[ 1 + 2 sin x + sin^2 x +1 - sin^2 x ] / common denom

[ 2 + 2 sin x ] / [(1+sin x) cos x ]


I think you can take it from there

Answer 2

for your second question you probably meant (I will switch ß to x, easier to type)
cosß- cosß/(1-tanß) = sinßcosß/(sinß-cosß)

Left Side
= cosx - cosx/(1-sinx/cosx)
= cosx - cosx/[(cosx-sinx)/cosx]
=cosx - cos^2 x/(cosx-sinx)
=(cos^2 x - sinxcosx - cos^2 x)/cosx - sinx)
= -sinxcosx/(cosx-sinx)
= sinxcosx/(sinx-cosx)
= Right Side