Hi How do I prove 2 sec^2 y + tan^2 y = 3 sec^2 y - 1? Thanks?

tan^2 y = sec^2 y - 1

2 sec^2 y + tan^2 y
= 2 sec^2 y + sec^2 y - 1
= 3 sec^2 y - 1

Method 1 : Use the identity sec²θ = tan²θ + 1, i.e. tan²θ = sec²θ - 1

L.H.S.
= 2 sec²y + tan²y
= 2 sec²y + (sec²y - 1)
= 2 sec²y + sec²y - 1
= 3 sec²y - 1
= R.H.S.

Hence, 2 sec²y + tan²y = 3 sec²y - 1


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Method 2 : Use the identity sin²θ + cos²θ = 1, i.e. 1 - sin²θ = cos²θ

L.H.S.
= 2 sec²y + tan²y
= 3 sec²y - sec²y + tan²y
= 3 sec²y - (1/cos²y) + (sin²y/cos²y)
= 3 sec²y - [(1 - sin²y)/cos²y)]
= 3 sec²y - (cos²y/cos²y)
= = 3 sec²y - 1
= R.H.S.

Hence, 2 sec²y + tan²y = 3 sec²y - 1

L.H.S=R.H.S

2sec^2y + tan^2y =
2sec^2y + sec^2y - 1 =
3sec^2y - 1