Question 122288
basically what you are trying to do here is prove that the left hand side and the right hand side are the same thing.  You use trig identities to do this.<br>

NOTE: for this to work out neatly I had to use the formula part of this site and it didn't let me do things with your notation so you need to know that:<br>

tan^2 x = (tan x)^2
sin^2 x = (sin x)^2
cos^2 x = (cos x)^2
cot^2 x = (cot x)^2<br>

1)  
{{{sec x (csc x - cot x * cos x) = tan x }}} --> Change cot x into cos x / sin x
{{{sec x(csc x - ((cos x)/(sin x) * cos x)) = tan x }}} --> get a common denominator
{{{sec x((1/sin x) - ((cos x)^2/(sin x))) = tan x }}} --> subtract 
{{{sec x((1-(cos x)^2)/(sin x)) = tan x }}} --> 1 - cos^2 x = sin^2 x
{{{sec x((sin x)^2/(sin x)) = tan x }}} --> divide
{{{sec x(sin x) = tan x }}} --> sec x = 1/cos x
{{{(1/cos x) *(sin x) = tan x }}} --> multiply
{{{ sinx / cos x = tan x }}} --> simplify
{{{ tan x = tan x }}}<br>

2)  
{{{(cos x)^2 + (tan x)^2 * (cos x)^2 = 1 }}} -->factor out a cos^2 x
{{{(cos x)^2 *(1 + (tan x)^2) = 1 }}} --> 1 + tan^2 x = sec^2 x
{{{(cos x)^2 *(sec x^2) = 1 }}} --> sec ^2 x = 1 / cos^2 x
{{{(cos x)^2 * 1/(cos x)^2 = 1 }}} --> multiply and simplify
{{{ 1 = 1 }}}<br>

I don't know how to do the third one