1. ## Trig Identity Help

One question left and i got stuck on it, i am starting to get the hang of these things though.

Thanks, joe

sec x + csc x / 1 + tan x = cscx

2. Originally Posted by DaCoo911
One question left and i got stuck on it, i am starting to get the hang of these things though.

Thanks, joe

sec x + csc x / 1 + tan x = cscx
Hi DaCoo911,

Working with the left hand side,

$\dfrac{\sec x + \csc x}{1+\tan x}=$

$\dfrac{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}{1+\dfrac{\sin x}{\cos x}}=$

$\dfrac{\dfrac{\sin x+ \cos x}{\sin x \cos x}}{1+\dfrac{\sin x}{\cos x}}=$

$\dfrac{\dfrac{\sin x+ \cos x}{\sin x \cos x}}{\dfrac{\cos x+\sin x}{\cos x}}=$

$\dfrac{\sin x+ \cos x}{\sin x \cos x}\cdot \dfrac{\cos x}{\sin x+\cos x}=$

$\dfrac{1}{\sin x}=$

$\csc x$

3. Dacoo,

1. $\frac{1}{cosx}=secx$

2. $\frac{1}{sinx}=csecx$

3. $\frac{sin}{cosx}=tgx$

Find common denominator and you'll have

$\frac{(sinx+cosx)/(cosxsinx)}{(cosx+sinx)/cosx}$

cancel $cosx+cosx$ then you got $\frac{cosx}{cosxsinx}$, cancel cosinuses and finally you got $1/sinx$ which is second identity i wrote.