Question

Verifying Identities:

How would I make this expression equal tanx?

sin^2x cosx
------ x -----
cosx 1

Answer 1

I know tangent is sin/cos but I can't seem to get the algebra right

Answer 2

0.0000000000000000000000000000001% sure that it is not 1

Answer 3

or you know you could cross multiply your duplicate and then divide what you have left

Answer 4

That's what I was thinking... here's the original identity:


tanx-sin(-x)
-----------
1 + cosx

Answer 5

dear neptune

Answer 6

Agreed

Answer 7

how do you even.....

Answer 8

sin 2x = 2 sinx cosx

Answer 9

sin^2 x = 1 - cos^2 x

Answer 10

what is the original problem?

Answer 11

\[\Large \frac{\tan(x)-\sin(-x)}{1+\cos(x)}\]
\[\Large \frac{\tan(x)+\sin(x)}{1+\cos(x)}\]
\[\Large \frac{\tan(x)+\sin(x)}{1+\cos(x)} * \frac{1-\cos(x)}{1-\cos(x)}\]
\[\Large \frac{\tan(x)+\sin(x)-\tan(x)*\cos(x)-\sin(x)*\cos(x)}{1-\cos^2(x)}\]
\[\Large \frac{\tan(x)+\sin(x)-\frac{\sin(x)}{\cos(x)}*\cos(x)-\sin(x)*\cos(x)}{\sin^2(x)}\]
\[\Large \frac{\tan(x)-\sin(x)*\cos(x)}{\sin^2(x)}\]
\[\Large \frac{\frac{\sin(x)}{\cos(x)}-\sin(x)*\cos(x)}{\sin^2(x)}\]
\[\Large \frac{\sin(x)\left(\frac{1}{\cos(x)}-\cos(x)\right)}{\sin^2(x)}\]
\[\Large \frac{\frac{1}{\cos(x)}-\cos(x)}{\sin(x)}\]
\[\Large \frac{\frac{\cos^2(x)-1}{\cos(x)}}{\sin(x)} = \frac{\cos^2(x)-1}{\cos(x)} * \frac{1}{\sin(x)}\]
\[\Large \frac{1-\sin^2(x)-1}{\cos(x)*\sin(x)} = \frac{\sin(x)*\sin(x)}{\cos(x)*\sin(x)} = \frac{\sin(x)}{\cos(x)} = \]

tan(x)

Answer 12

Did you follow?

Answer 13

if you change tan to sin/cos, the derivation is a bit shorter
\[ \frac{\tan x + \sin x}{1+\cos x}= \frac{1}{1+\cos x}\left( \tan x +\sin x\right) \\
= \frac{1}{1+\cos x}\left( \frac{\sin x}{\cos x} +\sin x\right)\\
= \frac{1}{1+\cos x}\left( \frac{\sin x}{\cos x} + \frac{\cos x\sin x}{\cos x}\right)\\
= \frac{1}{1+\cos x}\left( \frac{\sin x(1+\cos x)}{\cos x} \right)

\]
cancel the 1+cos x to get sin /cos = tan