1. Given,  1- (1- cos²T)      =  sin²T +1
                           tan²t

        1-     (sin ² T)               = sin²T + 1
                  sin²T/cos²T

         1-cos ²T  = 1+ sin²T

-Cos²T = sin²T (this is not identical)

2. So square both sides : Cos⁴T = Sin⁴T

3 . Cos⁴T - Sin⁴T =0

4. (cos²T-sin²T) (cos²2T+sin²T)=0

5. That gives you value of Theta as 45 degree. Am I right?

Where T=Theta.

I think the identity is wrong. it should be like this

(1-(1-cos² Theta)/tan²Theta)=sin²Theta

If u r trying to find the value of theta, no theta satisfies the above eqn. (except some complex angle..not sure if it does).

Parbatya is wrong, put 45 back to the equation, u will be ended up with 0.5=1.5

Actually this is not 'identity'. They can not be solved in straight way but they have solution as I found there.

BTW I am not mathematician.

LHS = 1- (1- cos²T) 
                     tan²t

        = 1-     (sin ² T)               ------->>> 1-cos²T = sin²T  &   tan ²T = sin²T/cos²T
                    sin²T/cos²T

        = 1-cos ²T

        = sin ² T------->>> 1-cos²T = sin²T

Therefore, RHS suppose to be----->>  sin ² T  !!  ?????  

Did I get it right !  WOW !  I better participate on the TV show " Are You Smarter Than a 5th Grader?". LOL ! :D :D :p

मिलेन मिलेन पर्बतेदा'। त्यो जहाँ पायो त्यैँ दुइतिरै स्क्वाएर घुसाउन मिल्ने भा' भे' त १ बराबर २ देखाउन नि सकिन्छ।

Here is simple solution: since theta is angle, I am not writing it at the very end. You may assume it.

1 = sin^2 + cost^2,

1/tan^2 = cos^2 / sin^2

1- cos^2 = sin^2

Use these expressions:

sin^2 + cos^2 + (1-cos^2)/tan^2 = 1 + sin^2

Cancel out common parts:

cos^2 - sin^2 * cos^2  / sin^2 = 1

cos^2 - cos^2 = 1

-cos^2 A = sin^2 A

Since there is -ve sign , and SinA and cosA are both squared term, it does not have real solution. It is April Fool problem.

1- 1- (sin ² T)             =   1-cos² T   कसरी हुन्छ?
      sin²T/cos²T

1- 1- (sin ² T)              
       sin²T/cos²T

‍= 1-1-cos²T

= cos²T  भएन र भन्या?

अनि सुरु देखि नै त्यो 1-1 किन बोकेर हिडि राखेको? त्यो त त्यसै 0 हुनुपर्ने।