establish this identity:

cosx + 1+sinx = 2 sec x
1+sinx cosx


establish this identity:

sec(squared)x - tan(squared)x + tan x = sin x + cos x
sec x

Thank you so much any help would be great!! :)

1.

\frac{cos(x)}{1+sin(x)} + \frac{1+sin(x)}{cos(x)} = 2 sec(x)

Make everything a single fraction:

\frac{cos^2(x)+(1+sin(x))^2}{(1+sin(x))(cos(x))}

Then simplify:

\frac{cos^2(x)+1+2sin(x)+sin^2(x)}{(1+sin(x))(cos( x))}

\frac{(cos^2(x)+sin^2(x))+1+2sin(x)}{(1+sin(x))(co s(x))}

\frac{1+1+2sin(x)}{(1+sin(x))(cos(x))}

\frac{2+2sin(x)}{(1+sin(x))(cos(x))}

\frac{2(1+1sin(x))}{(1+sin(x))(cos(x))}

\frac{2}{(cos(x))}

2sec(x)

2.

\frac{sec^2(x) - tan^2(x) + tan (x)}{sec(x)} = sin( x) + cos( x)

Use: tan^2(x) + 1 = sec^2(x)

\frac{(tan^2(x) +1) - tan^2(x) + tan (x)}{sec(x)}

\frac{tan^2(x) +1 - tan^2(x) + tan (x)}{sec(x)}

\frac{(tan^2(x) - tan^2(x))+1 + tan (x)}{sec(x)}

\frac{1 + tan (x)}{sec(x)}

Convert everything into cos and sin.

\frac{1 + \frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}}

\( 1 + \frac{sin(x)}{cos(x)}\) \times cos(x)

cos(x) + sin(x)

Hope all helped! :)

Yes thank you but how do I make each side equal to the other like 1=1 for example?

in fact, I should have typed :

\frac{cos(x)}{1+sin(x)} + \frac{1+sin(x)}{cos(x)} = \frac{cos^2(x)+(1+sin(x))^2}{(1+sin(x))(cos(x))}

Then continue with the simplifying. You just have to show that they are equal at the end, putting for example 'shown' or 'proved' at the end.

OK yes thank you very much again! Have a great day!

A 'rating' would be nice from you too! :)

OK I'm new at this today actually so I'm sorry but I don't know how to rate you! Could you help me out!

Just click on the 'rate this answer button, and follow what you are asked! :)