verifying trigonometric identities
2007-11-23 19:21:50 · 6 answers · asked by tobi 1 in Science & Mathematics ➔ Mathematics
1/sin²θ + 1/cos²θ
= (sin²θ+cos²θ)/[sin²θ cos²θ]
= 1/[sin²θ(1- sin²θ)]
= 1/ (sin²θ - sin ^4 θ)
2007-11-23 19:31:37 · answer #1 · answered by sahsjing 7 · 2⤊ 0⤋
1/sin²θ + 1/cos²θ=[sin²θ+cos²θ]/sin²θ cos²θ
=1/sin²θ cos²θ
=1/[sin²θ(1-sin²θ)]
=1/ [sin²θ - sin ^4 θ]
2007-11-24 04:20:05 · answer #2 · answered by kami 5 · 0⤊ 0⤋
LHS
(cos ² θ + sin ² θ) / (sin ² θ cos ² θ)
1 / [ (sin ² θ) (cos ² θ) ]
RHS
1 / [sin ² θ (1 - sin ² θ) ]
1 / [ (sin ² θ)(cos ² θ) ]
LHS = RHS
2007-11-23 21:11:51 · answer #3 · answered by Como 7 · 1⤊ 1⤋
1) cscx / (cotx + tanx) = (1/sinx) / [(cosx/sinx) + (sinx/cosx)] = (1/sinx) / [(cos²x + sin²x)/sinx cosx] = (1/sinx) / (1/sinx cosx) = (1/sinx)*(sinx cosx) = cosx. 2) [1/(siny -1)] - [1/(siny+1)] = [(siny+1) - (siny-1)] / (sin²y - 1) = (siny+1-siny+1) / -cos²y = 2*(1 / -cos²y) = 2*(-sec²y) =-2sec²y. 3) sin³x - cos³x = (sinx - cosx)(sin²x + sinx cosx + cos²x), by using the formula for difference of two cubes = (sinx - cosx)(sin²x + cos²x + sinx cosx) = (sinx - cosx)(1 + sinx cosx). 4) tanθ + [cosθ / (1 + sinθ)] = (sinθ/cosθ) + [cosθ / (1 + sinθ)] = [sinθ(1+sinθ) + cos²θ] / [cosθ(1 + sinθ)] = (sinθ + sin²θ + cos²θ) / [cosθ(1 + sinθ)] = (sinθ + 1) / [cosθ(1 + sinθ)] = 1 / cosθ = secθ.
2016-04-05 06:04:08 · answer #4 · answered by Donna 4 · 0⤊ 0⤋
87
2007-11-23 19:30:25 · answer #5 · answered by js2_26 2 · 0⤊ 2⤋
Sure! But only if you'll change my oil.
2007-11-23 19:31:56 · answer #6 · answered by tharnpfeffa 6 · 1⤊ 2⤋