# How to prove this trigonometric expression?

How would you go about proving the following?

$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A$$

This is what I've done so far:

$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$

....no idea how to proceed .... X_X

• It looks to me like you rewrote $\sin^2 A$ as $1+\cos^2A$. It should be $1-\cos^2A$. And that seems to be the end of your difficulty? Commented May 8, 2013 at 14:21
• oh ya. Thanks, I've corrected it. And unfortunately, no. Commented May 8, 2013 at 14:23
• Now simplify the numerator! Commented May 8, 2013 at 14:24
• You have already done it. cut out $\cos^2 x$ and u r done
– user67773
Commented May 8, 2013 at 14:26

You did everything thus far correctly, I just pick up with where you left off in the second line:

\begin{align}(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A) & = \dfrac{1 - 2 \cos A + \cos^2 A + \sin^2 A}{\sin A(1 - \cos A)} \\ \\ & = {1 \color{blue}{\bf + \cos^2 A} -2\cos A + 1 \color{blue}{\bf - \cos^2A} \over \sin A(1-\cos A)} \\ \\ & = \dfrac{2 - 2\cos A}{\sin A(1 - \cos A)}\\ \\ & = \dfrac{2\color{red}{\bf (1-\cos A)}}{\sin A\color{red}{\bf (1 - \cos A)}}\\ \\ & = \frac{2}{\sin A} \\ \\ & = 2 \csc A \end{align}

• if we express $\sin^2A=1-\cos^2A$ at your starting expression, we can safely cancel $1-\cos A$ straight away from the numerator & the denominator Commented May 8, 2013 at 15:39
• @amWhy: looks good to me and the color is great! +1 Commented May 9, 2013 at 0:29

$$LHS =\frac {1 - \cos A} {\sin A} + \frac {\sin A} {1 - \cos A}$$ $$= \frac {2 \sin^2 \frac A2} {2\sin \frac A2 \cos \frac A2} + \frac {2\sin \frac A2 \cos \frac A2}{2 \sin^2 \frac A2}$$

$$= \frac {\sin \frac A2} {\cos \frac A2} + \frac {\cos \frac A2} {\sin \frac A2}$$

Now just cross multiply and you get the answer.

Hint:

$1-\cos A=1-(1-2 \sin^2\dfrac{A}{2})=2\sin^2 \dfrac{A}{2}$

$\dfrac{2 \sin^2 \dfrac{A}{2}}{2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}}=\tan \dfrac{A}{2}$

The other expression will be $\cot \dfrac{A}{2}$

$(\tan^2 \dfrac{A}{2}+1) /\tan\dfrac{A}{2}= \dfrac{\sec^2 A \cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}= \dfrac{1}{2 \sin A}$

I'll go step by step.

$${1 - \cos A \over \sin A} + {\sin A \over 1 - \cos A}$$

$${(1 - \cos A)^2 \over \sin A (1 - \cos A)} + {\sin^2 A \over \sin A(1 - \cos A)}$$

$$(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A)$$

Expanding $(1 - \cos A)^2$ yields:

$$1 - 2\cos A + \cos^2 A + \sin^2 A \over \sin A(1 - \cos A)$$

Knowing that $\cos^2 x + \sin^2 x = 1$, we can change the expression to the following:

$$1 - 2 \cos A + 1 \over \sin A(1 - \cos A)$$

$$2 - 2 \cos A \over \sin A(1 - \cos A)$$

We factorize using $2$:

$$2(1 - \cos A) \over \sin A(1 - \cos A)$$

$$2 \over \sin A$$

$$2 \left ({1 \over \sin A} \right)$$

Finally:

$$2 \csc A$$

• Thanks a lot! I greatly appreciate all the effort :D Commented May 8, 2013 at 14:39

When we "cross-multiply," on top we get $(1-\cos A)^2+\sin^2 A$. Expand the square.

We get $1-2\cos A+\cos^2 A+\sin^2 A$. Replace $\cos^2 A+\sin^2 A$ by $1$. We get $2-2\cos A$ on top, and now it's over.

LCM is not required.

Observe that the first term already has $\sin A$ in the denominator.

$$\text{Now,}\frac{\sin A}{1-\cos A}=\frac{\sin A(1+\cos A)}{1-\cos^2A}=\frac{\sin A(1+\cos A)}{\sin^2A}=\frac{1+\cos A}{\sin A}$$

$$\text{So,}{1- \cos A \over \sin A } + { \sin A \over 1- \cos A} =\frac{1-\cos A}{\sin A}+\frac{1+\cos A}{\sin A}=\frac2{\sin A}=2\csc A$$