Simple pre calc question?

Question

How do i prove that both sides are equal of the given identity for the given vaules of the variable t.

tan^2t + 1 = sec^2 t

a) t= 3pie/4

b)t= -2pie/3

i seem to be doing something wrong... or maybe because its 3am :(..

I know it is.. but the problem is. i need to show work :(...

Answer 1

tan^2t+1 = sec^2t
(sin^2t)/(cos^2t)+1 = 1/(cos^2t)
(sint)^2/(cost)^2+ 1 = 1/(cost)^2
sin 3pie/4 = -1/squareroot (2)
cos 3pie/4 = 1/squareroot (2)

Put them in above equation. After simplifying u will get

(-1)^2 + 1 = 1 / (1/squareroot (2))^2
(-1)^2 + 1 = (squareroot (2))^2
2 = 2

second one is exectly same. just put down values of sin -2pie/3 and cos -2pie/3

Answer 2

tan² t + 1 = sec^2 t . . . this is already simplified Pythagorean
the expression on left and right will just cancel
you can not solve t
a). . . t= 3pie/4 . . . . this is just a constant . . .no need to prove

b). . .t= -2pie/3 . . . . this is just a constant . . .no need to prove

Answer 3

a)
tan^2(3π/4) + 1 =? sec^2(3π/4)
(- 1)^2 + 1 =? (√2)^2
1 + 1 = 2

b)
tan^2(- 2π/3) + 1 =? sec^2(- 2π/3)
(√3)^2 + 1 =? (-2)^2
3 + 1 = 4