Question

Simplify

{cos (theta) / 1-sin (theta) } - { cos(theta) / 1+ sin(theta) }

Answer 1

\[\frac{\cos (\theta) }{ 1-\sin (\theta) } - \frac{ \cos(\theta) }{ 1+ \sin(\theta) }\]

Answer 2

subtract !

Answer 3

\[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\]

Answer 4

But the denominator is not the same !! is it ?? !!

Answer 5

they are not the same, that doesn't mean you cannot subtract. it means you have to use the method for subtracting fractions i wrote above

Answer 6

\[\frac{\cos (x) }{ 1-\sin (x) } -\frac { \cos(x) }{ 1+ \sin(x) }\]
\[=\frac{\cos(x)(1+\sin(x))-\cos(x)(1-\sin(x))}{(1+\sin(x))(1-\sin(x))}\] then some algebra and finally some trig identity

Answer 7

ok , thank you

Answer 8

I stuck here !!

\[\frac{ \cos \theta + \cos \theta \times \sin \theta}{ 1 - \sin ^{2}\theta} -\frac{ \cos \theta - \cos \theta \times \sin \theta }{ 1- \sin ^{2} \theta }\]

What Next ?! ?! ?!

Answer 9

I tried and this what I got ...

\[\frac{ 1 + \sin \theta }{ \cos}\]

then ?

Answer 10

* I mean cos (theta)

Answer 11

\[\dfrac{\cos\theta }{1 - \sin\theta} - \dfrac{\cos\theta}{1 + \sin\theta}\]Let's first try to get the same denominator. Try multiplying the first fraction by \(\large \frac{1 + \sin\theta}{1 - \sin \theta}\) and the second fraction by \(\large \frac{1- \sin\theta}{1 - \sin\theta}\).

Answer 12

Ok , I did that ...
next

Answer 13

\[= \dfrac{\cos\theta ( 1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)} - \dfrac{\cos\theta ( 1- \sin\theta)}{( 1 + \sin\theta ) (1 - \sin\theta)}\]\[= \dfrac{\cos \theta + \sin \theta \cos \theta - \cos\theta + \sin\theta \cos \theta }{( 1 - \sin\theta)(1 + \sin\theta)}\]\[= \dfrac{2\sin\theta \cos \theta}{ 1 - \sin^2 \theta}\]\[= \dfrac{2\sin\theta\cos\theta}{\cos^2\theta}\]\[=2 \dfrac{\sin\theta}{\cos\theta}\]\[= 2\tan \theta\]

Answer 14

why 2 ??

Answer 15

\[\cos\theta + \sin\theta \cos \theta - \cos\theta + \sin\theta \cos \theta = 2\sin\theta\cos\theta\]

Answer 16

Thanks