Trigonometric Identities cosecA

(Extract from Cambridge O-Level Additional Mathematics June 2012 exam paper 1, question 3)

Show that

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

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

Help please.

Re: Trigonometric Identities cosecA

Quote:

Originally Posted by

**zikcau25** Show that

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

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

Help please.

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

$\displaystyle \frac{cos(A)}{sin(A)} + \frac{sin(A)}{1 + cos(A)} = \frac{1}{sin(A)}$

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

So what's the LHS look like?

-Dan

Re: Trigonometric Identities cosecA

Another approach:

Try to derive the following:

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

$\displaystyle \tan\left(\frac{A}{2} \right)=\frac{\sin(A)}{1+\cos(A)}$

Re: Trigonometric Identities cosecA

Hello, zikcau25!

Quote:

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

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

. . . . . . $\displaystyle =\;\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} $

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