Identity crisis

**nightrider456** · January 11th 2010, 06:54 PM

Heres the problem, Prove the identity. I can't for the life of me to figure it out.

Cosx+sinxtanx = cscx
sinxsecx

**Amer** · January 11th 2010, 07:16 PM

Originally Posted by nightrider456

Heres the problem, Prove the identity. I can't for the life of me to figure it out.

Cosx+sinxtanx = cscx
sinxsecx

$\dfrac{\cos x + \sin x \tan x }{\sin x \sec x}$

$\frac{\cos x + \sin x \cdot \dfrac{\sin x}{\cos x}}{\sin x \dfrac{1}{\cos x }}$

$\dfrac{\dfrac{\cos ^2 x + \sin ^2 x}{\cos x}}{\dfrac{\sin x}{\cos x}}$

$\frac{\cos ^2 x + \sin ^2 x }{\sin x } = \frac{1}{\sin x} = \csc x$

**Prove It** · January 11th 2010, 07:18 PM

Originally Posted by nightrider456

Heres the problem, Prove the identity. I can't for the life of me to figure it out.

Cosx+sinxtanx = cscx
sinxsecx

$\frac{\cos{x} + \sin{x}\tan{x}}{\sin{x}\sec{x}} = \frac{\cos{x}}{\sin{x}\sec{x}} + \frac{\sin{x}\tan{x}}{\sin{x}\sec{x}}$

$= \frac{\cos^2{x}}{\sin{x}} + \frac{\tan{x}}{\sec{x}}$

$= \frac{\cos^2{x}}{\sin{x}} + \frac{\frac{\sin{x}}{\cos{x}}}{\frac{1}{\cos{x}}}$

$= \frac{\cos^2{x}}{\sin{x}} + \sin{x}$

$= \frac{\cos^2{x}}{\sin{x}} + \frac{\sin^2{x}}{\sin{x}}$

$= \frac{\cos^2{x} + \sin^2{x}}{\sin{x}}$

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

$= \csc{x}$ .

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