Question

prove that
sinA - 2sin²A/2cos²A - cosA = tanA​

Answer 1

Step-by-step explanation:

Muntiplying 0 both side

{sinA - 2sin²A/2cos²A - cosA}0= 0{tanA}

0 = 0 Hence proved

Answer 2

ur question is wrong , it's not square it's cube.

if it is cube then the answer is,

(sinA-2sin³A)/(2cos³A-cosA)

= sinA(1-2sin²A)/(cosA(2cos²A-1))

= sinAcos(2A)/(cosAcos(2A))

= (sinA/cosA)

= tanA

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