Question

Prove \[\sin 2\theta \over \cos \theta \] + \[\cos 2\theta \over \sin \theta\] = \[\csc \theta\]

Answer 1

2 Sin (x)Cos(x) Cos^2(x)-Sin^2(x)
----------- + ------------------
Cos(x) sin(x)


2 Sin (x)Cos(x) (1- Sin^2(x))-Sin^2(x)
----------- + ------------------
Cos(x) sin(x)

2 Sin(x) +1-2 sin(x)
-------------------
sin(x)


1
----- = csc(x)
sin(x)

Answer 2

I don't get what has happened in between 2nd line to the 3rd..?

Answer 3

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

Answer 4

what I did in second line
2 Sin (x)Cos(x)
-----------
Cos(x)

cos(x) cancels

2 sin(x)

Answer 5

You guys are awesome! but I still am abit connfused :s \[2\sin \theta \cos \theta \over \cos \theta \] can become \[2\sin \theta\] times \[\sin \theta \over \sin \theta \]
?

Answer 6

oh wow! imranmeah! thanks i didn't catch cancelling cos :p I got it now!! Thanks alot.