Question

(WILL GIVE MEDAL AND FAN)
(I'll post the equations below)
Simplify.

Answer 1

\[\tan x (\sin x + \cot x \cos x)\]

Answer 2

\[\frac{ 1 }{ \sec^2x } + \frac{ 1 }{ \csc^2x }\]

Answer 3

tan x (sin x + cot x cos x)
= (sin x / cos x) * (sin x + (cos x / sin x) * cos x)
= (sin x / cos x) * (sin x + ((cos x)^2 / sin x))
= (sin x / cos x) * (((sin x)^2 / sin x) + ((cos x)^2 / sin x))
= (sin x / cos x) * ((sin x)^2 + (cos x)^2) / sin x)
= (sin x / cos x) * (1 / sin x)
= 1 / cos x
= sec x

Answer 4

Ah, so that's how you do it. What about the second one? ☺

Answer 5

Give me a chance to sole that one... :D

Answer 6

Alrighty :)

Answer 7

Recall that 1/sec²(x) = cos²(x), 1/csc²(x) and sin²(x) + cos²(x) = 1.

This yields:

cos²(x) + sin²(x)
==> 1

Answer 8

I see! I never really do get this stuff, haha
What about \[\sec x - \sin x \tan x\]

Answer 9

1/cosx- sinx·sinx/cosx= 1/cosx - sin²x/cosx =(1-sin²x)/cosx=
cos²x/cosx = cosx

Can u post in new question? Thx...

Answer 10

Sure!