Question

trig identity

Answer 1

\[\large \frac{ 1+2sinxcosx }{ \cos^2x-\sin^2x }\]

Answer 2

Which part did you get stuck at

Answer 3

that equals \[\large\frac{ cosx+sinx }{ cosx-sinx }\]

Answer 4

i can't seem to make them equal :(

Answer 5

1 + 2sinx cosx = sin²x + 2sinxcosx + cos²x = ( sinx + cosx)²

Answer 6

cos²x - sin²x = ( cosx - sinx ) ( cosx + sinx)

Answer 7

Now simplify upper and lower => the left side :)

Answer 8

this right

Answer 9

Can you simplify them?

Answer 10

yup, got it cosx+sinx/cosx-sinx

Answer 11

LHS=RSH

thankyou ! :D