cos4thita/cos2phi+ sin 4 thita/sin2phi=1 given to prove cos4phi/cos2thita+ sin 4 phi/sin2thita plz if u provide online chating with tutor u...

We'll re-write the condition, multiplying both sides by
the least common denominator, that is the product of denominators of the terms from the
left side: cos2pi*sin 2pi.

But cos 2pi = 1 and sin 2pi = 0
=> cos2pi*sin 2pi = 1*0 = 0

cos 4theta/cos2pi+ sin 4
theta/sin2pi=1

(cos2pi*sin 2pi*cos 4theta)/(cos2pi)+
(cos2pi*sin 2pi*sin 4 theta)/(sin2pi)=1*cos2pi*sin
2pi

We'll reduce like
terms:

sin 2pi*cos 4theta+ cos2pi*sin 4 theta
=1*0

sin (2pi + 4theta) =
0

2pi + 4theta = arcsin 0

2pi
+ 4theta = 0

4theta =
-2pi

theta = -2pi/4

theta =
-pi/2

Under this conditions, we'll calculate the
expression, having theta= -pi/2.

cos4pi/cos2*(-pi/2)+ sin 4
pi/sin2*(-pi/2)

cos -pi = cos pi =
-1

sin -pi = -sin pi = 0

-1/-1
+ 0/0

Since the second fraction is
meaningless (we cannot divide by 0), the expression cannot be computed under this
conditions.