Question

More proving trig
\(\bf {\dfrac{sin~x}{1~-~cos~x}+\dfrac{sin~x}{1~+~cos~x}~=~2~csc~x}\\\)

Answer 1

factor out sinx and add up the fractions

Answer 2

I know that csc=1/sin but I'm not sure if I need that here or not.

Answer 3

\(\bf sin(x)(\dfrac{1+1}{1-cos(x)+1+cos(x)})\)?

Answer 4

you should get

\[\bf sin x\left(\dfrac{1+1}{(1-cos x)(1+cos x)}\right)\]

Answer 5

next recall the identity : \((a-b)(a+b)=a^2-b^2\)

Answer 6

you need to have common denominator to add fractions ^

Answer 7

can I continue the answer ?

Answer 8

I know to solve this

Answer 9

Which gives \(\bf sinx\left(\dfrac{1+1}{1^2-cosx^2}\right)\)

Answer 10

sure, please go ahead :)

Answer 11

you can have two fractions in the left have common denominator

Answer 12

(1-cos^2(x));

if we made this you will get in left side :-
(sin(x)*(1-cos(x)) + sin(x)*(1+cos(x)))/1-cos^(x)

(sinx + sin(x)*cos(x) + sin(x) -sin(x)*cos(x) )/ (1- cos^(X)) // sinx*cos(X) gone !!!
2*sin(x)/1-cos^2(x) // ut 1= sin^(x) + cos^(x)
2sin(x)/sin^(x) = 2/sin(x) = 2/csc(x)

Answer 13

then we proof the left side equal to right side :D