Question

i don't know how to solve this, i'm so lost.

determine the exact value of c

cot( 4c- pie/4) + tan ( 2c + pie/4)= 0

please help me!!!

THANK YIU

Answer 1

cot(4c-π/4) = 1/tan(4c-π/4)
= 1/[ (tan4c) - tanπ/4)/(1 + tan4c(tanπ/4))]
= (1+tan4ctanπ/4)/(tan4c - tanπ/4)
= (1 + tan4c)/(tan4c - 1) , since tan π/4 = 1

also from tan 2A = 2tanA/(1 - tan^2 A)
the above is
= [ 1 + 2tan2c/(1 - tan^2 2c) ] / [2tan2c/(1 - tan^2 2c) - 1 ] ....ARGGGG!
let tan 2c = x
then above
= ([ 1 + 2x/(1-x^2) ] / [2x/(1-x^2) - 1 ]
= (1-x^2+2x)/(2x-1+x^2)

similarly tan(2c + /4)
= (tan2c + tan π/4) / (1 - tan2ctanP/4)
= (x+1)/(1-x)

original equation becomes
(1-x^2+2x)/(2x-1+x^2) = -(x+1)/(1-x)
cross-multiply .... and simplify

2x^3 + 2x = 0
2x(x^2 + 1) = 0
x = 0 or x^2+1=0
no solution for the second part, so

x = 0
tan 2c = 0
2c = 0 , 180, 360 , 540, 720..

c = 0, 90, 180, 270 , 360 , ... (degrees)

(somehow I think there is an easier way)