SOLUTION: I need help with these, I can't seem to figure them out (sec x-tan x)(sin x+1)= cos x 1-sin^2x = cosx/ tanx+cotx tan^2 x - 1/ tan^2 x + 1 = 1 - 2 cos^2 x

Algebra ->  Trigonometry-basics -> SOLUTION: I need help with these, I can't seem to figure them out (sec x-tan x)(sin x+1)= cos x 1-sin^2x = cosx/ tanx+cotx tan^2 x - 1/ tan^2 x + 1 = 1 - 2 cos^2 x      Log On


   

Question 930153: I need help with these, I can't seem to figure them out
(sec x-tan x)(sin x+1)= cos x
1-sin^2x = cosx/ tanx+cotx
tan^2 x - 1/ tan^2 x + 1 = 1 - 2 cos^2 x
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

1.
%28sec%28+x%29-tan%28+x%29%29%28sin+%28x%29%2B1%29=+cos+%28x%29
start with

%28sec%28+x%29-tan%28+x%29%29%28sin+%28x%29%2B1%29..............use identities sec%28+x%29=1%2Fcos%28x%29 and tan%28+x%29%28sin%28+x%29%2Fcos%28x%29%29

=%281%2Fcos%28x%29+-sin%28+x%29%2Fcos%28x%29%29%28sin+%28x%29%2B1%29

=%281%2Fcos%28x%29+-sin%28+x%29%2Fcos%28x%29%29%28sin+%28x%29%2B1%29

=%28%281-sin%28+x%29%29%28sin+%28x%29%2B1%29%29%2Fcos%28x%29

=%281%5E2-sin%5E2%28+x%29%29%2Fcos%28x%29

=%281-sin%5E2%28+x%29%29%2Fcos%28x%29 .......use identity %281-sin%5E2%28+x%29%29=cos%5E2%28x%29

=cos%5E2%28x%29+%2Fcos%28x%29

=cos%5Ecross%282%29%28x%29+%2Fcross%28cos%28x%29%29

=cos%28x%29+



2.
1-sin%5E2%28x%29+=+cos%28x%29%2F+%28tan%28x%29%2Bcot%28x%29%29...I am not quite sure what you need here, this is not true statement


3.
%28tan%5E2+%28x+%29-+1%29%2F%28+tan%5E2%28+x%29+%2B+1%29+=+1+-+2+cos%5E2%28+x%29
start with
%28tan%5E2+%28x%29-1%29%2F%28tan%5E2%28x%29%2B+1%29

=%28sin%5E2%28x%29%2Fcos%5E2%28x%29+-1%29%2F%28+sin%5E2%28x%29%2Fcos%5E2%28x%29+%2B+1%29+

=

=

=

=%28%28sin%5E2%28x%29+-+cos%5E2%28x%29%29%29%2F%28+%28sin%5E2+%28x%29%2B+cos%5E2%28x%29%29%29...sin%5E2%28x%29=1-cos%5E2%28x%29

=

=%28%281-2cos%5E2%28x%29%29%2F1%29

=1-2cos%5E2%28x%29