# Trigonometric Identities cosecA

• Nov 6th 2012, 05:36 PM
zikcau25
Trigonometric Identities cosecA
(Extract from Cambridge O-Level Additional Mathematics June 2012 exam paper 1, question 3)

Show that

$cotA+\frac{sinA}{1+cosA}\equiv cosecA$

I can't get rid of the 1+cosA when simplifying RHS, it keeps getting complex.
• Nov 6th 2012, 05:52 PM
topsquark
Re: Trigonometric Identities cosecA
Quote:

Originally Posted by zikcau25
Show that
$cotA+\frac{sinA}{1+cosA}\equiv cosecA$

I can't get rid of the 1+cosA when simplifying RHS, it keeps getting complex.

There can be easier solutions to the problem than the way I do it, but usually it works out not too bad.

I always replace things like cot(A) and cosec(A) (and whatever else) with sines and cosines. So for your identity we have
$\frac{cos(A)}{sin(A)} + \frac{sin(A)}{1 + cos(A)} = \frac{1}{sin(A)}$

Generally the next step is to add fractions. So we have fractions...let's add them. The common denominator will simply be $sin(A) \cdot (1 + cos(A))}$

So what's the LHS look like?

-Dan
• Nov 6th 2012, 06:20 PM
MarkFL
Re: Trigonometric Identities cosecA
Another approach:

Try to derive the following:

$\tan\left(\frac{A}{2} \right)=\csc(A)-\cot(A)$

$\tan\left(\frac{A}{2} \right)=\frac{\sin(A)}{1+\cos(A)}$
• Nov 6th 2012, 06:45 PM
Soroban
Re: Trigonometric Identities cosecA
Hello, zikcau25!

Quote:

$\text{Show that : }\:\cot A+\frac{\sin A}{1+\cos A}\:=\:\csc A$

We have: . $\frac{\cos A}{\sin A} + \frac{\sin A}{1 + \cos A} \;=\;\frac{\cos A}{\sin A} + \frac{\sin A}{1 + \cos A}\cdot\color{blue}{\frac{1-\cos A}{1-\cos A}}$

. . . . . . $=\;\frac{\cos A}{\sin A} + \frac{\sin A(1-\cos A)}{1-\cos^2\!A} \;=\;\frac{\cos A}{\sin A} + \frac{\sin A(1-\cos A)}{\sin^2\!A}$

. . . . . . $=\;\frac{\cos A}{\sin A} + \frac{1-\cos A}{\sin A} \;=\;\frac{\cos A + 1 - \cos A}{\sin A} \;=\;\frac{1}{\sin A} \;=\;\csc A$