cosA1sinA1sinAcosASolutionHope you want this expression to b

[(cosA)/(1+sinA)]+[(1+sinA)/(cosA)]

Solution

Hope you want this expression to be simplified .

Simplification:

[(cosA)/(1+sinA)+[(1+sinA)/cosA)]

There are two terms with different denominators. We write equivalent expressions by making a common denominator with the LCM, (1+sinA)cosA, of the denominators:

[(cosA)^2+(1+sinA)^2]/[(1+sinA)(cosA)]

Rewrite the above expression, expanding (1+sinA)^2 =1+2sinA+(sinA)^2:

[(cosA)^2+1+2sinA+(sinA)^2]/[(1+sinA)(cosA)]

Rearrange the numerator:

[(cosA)^2+(sinA)^2+1+2sinA]/[(1+sinA)(cosA)]

(cosA)^2+(sinA)^2=1 is a trigometic identity. So we can replace (cosA)^2+(sinA)^2 by 1:

[1+1+2sinA]/[(1+sinA)(cosA)]

=[2+2sinA)/[1+sinA)cosA]

=2(1+sinA)/[(1+sinA)cosa)]

=2/cosA

=2secA. Therefore,

[cosA/(1+sinA)]+ [(1+sinA)/cosA] = 2secA or 2/cosA

Hope this helps.