Question

Prove the trig. identity:
((sinx)/(1+cosx))+((1+cosx)/(sinx))=2cscx

Answer 1

best bet is to add. if it is not clear how to add this stuff up, replace
sin(x) by a and cos(x) by b and add
\[\frac{b}{1+a}+\frac{a}{1+b}\]

Answer 2

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

Answer 3

the race is on!

Answer 4

after you add, replace back. then i bet the only "trig" step will be to write
\[\sin^2(x)+\cos^2(x)=1\] and you will get your answre

Answer 5

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

Answer 6

lol @ joe

Answer 7

LHS=(2+2sincosx)/(1+cosx)sinx

Answer 8

i either win or lose, depending on whether you think i actually wrote an answer.
@kbix, most of these problems just require a bunch of algebra. don't let the "sine" and "cosine" fool you or make it more difficult. they are usually cooked up so that at some point you will see
\[\sin^2(x)+\cos^2(x)\] which you then replace by 1

Answer 9

just combine the two fractions:
(sinx / 1 + cosx) + (1 + cosx / sinx)
=sin^2(x)/ sin(x) *(1 +cos(x))^2/ sin(x) *(1 +cos(x))
=sin^2(x) +(1 +cos(x))^2/ sin(x) *(1 +cos(x))
=sin^2(x) +cos^2(x) +2cos(x) +1/ sin(x) *(1 +cos(x))
=1 +2cos(x) +1/ sin(x) *(1 +cos(x))
=2 +2cos(x) / sin(x) *(1 +cos(x))
=2 *(1 +cos(x)) / sin(x) *(1 +cos(x))
=2/ sin(x)
=2csc(x)

Answer 10

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

Answer 11

2/sinx

Answer 12

i feel cheated

Answer 13

\[\frac{b}{1+a}+\frac{1+a}{b}=\]
\[=\frac{b^2+(1+a)(1+a)}{(1+a)b}\]
\[=\frac{b^2+a^2+2a+1}{(1+a)b}\]

Answer 14

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

\[\frac{2+2\cos(x)}{(1+\cos(x))\sin(x)}\]
that is the ONLY trig step. the rest is algebra pure and (not so) simple

Answer 15

now factor out the 2, cancel the 1 + cos(x) and you get your answer.

Answer 16

@lagrange thanks. i am very slow at latex tonight

Answer 17

Thank-you soooo much, guys. I am pretty much awful at proofs. I greatlyyyyy appreciate it. (: