Proving Trig identities help!

**webb84** · March 28th 2011, 01:14 AM

Hi

having trouble with the below trig identity just cant seem to get round it

prove the trigonometric identity

sin^2A(secA+cosecA)/1+tanA = sinA

i admit not the best at trig, just not sure what to do with the bottom half any help would be appreciated

webb

**Prove It** · March 28th 2011, 01:19 AM

I assume this is your equation...

$\displaystyle \frac{\sin^2{A}(\sec{A} + \csc{A})}{1 + \tan{A}} \equiv \sin{A}$

Brackets MUST be used where they are needed...

Anyway...

$\displaystyle \frac{\sin^2{A}(\sec{A} + \csc{A})}{1 + \tan{A}} \equiv \frac{\sin^2{A}\left(\frac{1}{\cos{A}} + \frac{1}{\sin{A}}\right)}{1 + \frac{\sin{A}}{\cos{A}}}$

$\displaystyle \equiv \frac{\sin^2{A}\left(\frac{\sin{A} + \cos{A}}{\sin{A}\cos{A}}\right)}{\frac{\cos{A} + \sin{A}}{\cos{A}}}$

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

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

$\displaystyle \equiv \sin{A}$ .

**webb84** · March 28th 2011, 04:33 AM

Thankyou, that is exactly what I was after, the step by step helped brilliantly

much appreciated

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