Question

the value of the expression 1-(sin^2y/[1+cos y])+([1+cos y]/siny)-(sin y/[1-cos y])=????

Answer 1

can you use the equation editor

Answer 2

kkk

Answer 3

wait maybe i can make it out

Answer 4

...................

Answer 5

cos(y)

Answer 6

\[1-(\sin ^{2}y \div1+\cos y)+(1+\cos y \div \sin y)-(\sin y \div1-\cos y)\]

Answer 7

howwww??????????????????

Answer 8

give me just a second

Answer 9

is tht much step necessary????

Answer 10

i don't know
i was playing with

Answer 11

can u giv short steps???

Answer 12

i need quicker way

Answer 13

just combine fractions earlier instead of multiplying by conjugates

Answer 14

I got cos y

Answer 15

show

Answer 16

that pdf posted above is the complete solution, no shortcuts. Just remember that sin^2 y + cos^2 y = 1 &&& (1+cosy)(1-cosy) = (1-cos^2 y)

Answer 17

\[1-\frac{\sin^2(y)}{1+\cos(y)} \cdot \frac{1-\cos(y)}{1-\cos(y)}+\frac{(1+\cos(y))(1-\cos(y))-\sin(y)\sin(y)}{\sin(y)(1-\cos(y))}\]
\[1-\frac{\sin^2(y)(1-\cos(y))}{1-\cos^2(y)}+\frac{1-\cos^2(y)-\sin^2(y)}{\sin(y)(1-\cos(y))}\]
\[1-\frac{\sin^2(y)(1-\cos(y))}{\sin^2(y)}+\frac{\sin^2(y)-\sin^2(y)}{\sin(y)(1-\cos(y))}\]
\[1-(1-\cos(y))+\frac{0}{\sin(y)(1-\cos(y))}\]
\[1-1+\cos(y)+0=\cos(y)\]

Answer 18

Don't forget that y cannot be 180 degrees or 0 degrees!

Answer 19

\[\sin(y) \neq0 or 1-\cos(y) \neq 0\]
\[y \neq n \pi, n \in \mathbb{Z} \]

Answer 20

^ Or you can use overly complicated notation :D Great job ininaya :D

Answer 21

lol

Answer 22

Is there a way I can delete my medal? I don't deserve it :D

Answer 23

well i was thinking about saying when that bottom is not zero but you said it for me
when i didn't feel like it
so you deserve it because what you said is true

Answer 24

You are awesome. PERIOD (not monthly discharge).