Question

1−sin2y/(1+cosy)+(1+cosy)/siny+siny/(1−cosy)

Answer 1

1-1-cos^2y/cosy+1-cosy/siny-siny/1-cosy

Answer 2

1-(1-cosy)(1+cosy)/cosy+1+cosy/siny+siny/1-cosy

Answer 3

\[1-\sin^2y/(1+cosy)+(1+cosy)/siny+siny/(1-cosy)\]

Answer 4

@ganeshie8

Answer 5

1-(1-cosy)+(1+cosy)/siny+siny/(1-cosy)

Answer 6

question is not clear,
could you pelase use latex..

Answer 7

to find the value of the expression

Answer 8

3rd one

Answer 9

\[\large \dfrac{1-\sin^2y}{1+\cos y}+\dfrac{1+cosy}{\sin y}+\dfrac{\sin y}{1-\cos y}\]

like this ?

Answer 10

but 1- (sin^2y)/(1+cosy)

Answer 11

\[\large 1-\dfrac{\sin^2y}{1+\cos y}+\dfrac{1+cosy}{\sin y}+\dfrac{\sin y}{1-\cos y}\]

like this ?

Answer 12

yes

Answer 13

\[1-(1-cosy)+(1+cosy)/siny+siny/(1-cosy)\]

Answer 14

\[\large 1-\dfrac{\sin^2y}{1+\cos y}+\dfrac{1+cosy}{\sin y}+\dfrac{\sin y}{1-\cos y}\]


\[\large 1-\dfrac{1-\cos^2y}{1+\cos y}+\dfrac{1+cosy}{\sin y}+\dfrac{\sin y}{1-\cos y}\]

\[\large 1-\dfrac{(1+\cos y)(1-\cos y)}{1+\cos y}+\dfrac{1+cosy}{\sin y}+\dfrac{\sin y}{1-\cos y}\]

\[\large \cos y+\dfrac{1+cosy}{\sin y}+\dfrac{\sin y}{1-\cos y}\]

Answer 15

fine, so far ?

Answer 16

what was the given answer ?