Question

Please help!!!!

verify the identity.
(cosx)/(1+sinx) + (1+sinx)/(cosx)=2 sec x

Answer 1

That can be written as :

[cosx*cosx + (1+sinx)*(1+sinx) ] / [(1+sinx)*cosx]

[cosx*cosx + sinx*sinx +1 + 2sinx ]/[(1+sinx)*cosx]

[1 + 1 + 2sinx]/[(1+sinx)*cosx]

[2(1+sinx)]/[(1+sinx)*cosx]

=2/cosx

=2secx

Answer 2

you can do these more easily, since it is almost all algebra, by replacing \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\) and adding
\[\frac{a}{1+b}+\frac{b}{1+a}\]

Answer 3

actually it would be
\[\frac{a}{1+b}+\frac{1+b}{a}\]

Answer 4

but u still need to use sinx*sinx + cosx*cosx = 1

which algebra can't provide you. even if u represent using a and b, you will ultimately need trigonometry

Answer 5

then you get
\[\frac{a^2+(1+b)^2}{(1+b)a}\]
\[\frac{a^2+1+2b+b^2}{(1+b)a}\]
now we use the only small bit of trig, namely that \(a^2+b^2=1\) and get
\[\frac{2+2b}{(1+b)a}\]
\[\frac{2(1+b)}{(1+b)a}\]
\[\frac{2}{a}\]

Answer 6

@Saikam is right of course
you need one bit of trig
but often the algebra is so annoying writing all the sines and cosines confuses the issue
in this case the algebra wasn't so bad