Question

The first and correct answer will be marked as brainliest!!!

Answer 1

Answer:

I think this may help u further calculation u can do

Answer 2

Answer:

Step-by-step explanation:

I will assume and identity, so will work on the right side and simplify.  

sin2A/cos2A + cos2A/sin2A => (sin²2A + cos²2A)/(cos2Asin2A)  

Now, sin²θ + cos²θ ≡ 1 so,  

1/(cos2Asin2A).....(1)  

Also, sin2θ ≡ 2sinθcosθ  

sin4A = 2sin2Acos2A  

Hence, sin2Acos2A = (sin4A)/2  

Therefore, from (1) we get: 1/[(sin4A)/2)  

i.e. 2/sin4A => 2cosec4A....should be right side

HOPE THIS HELPS YOU... PLZ MARK THIS AS BRAINLIEST....