LHS (left hand-side)=csc(π-θ)=1/sin(π-θ) =1/cos(θ)=secθ=RHS sin(π-θ)=cos(θ) because sin(π-θ)=sin(π)sinθ-cos(π)cosθ =0×sinθ-(-cosθ)=0+cosθ=cosθ
2 Answers · Education & Reference · 07/04/2009
LHS = 1 + cos (2x) - cos² x = 1 + 2 cos² x - 1 - cos² x ... [using the double angle formula cos (2u) = 2 cos² u - 1] = cos² x = RHS
1 Answers · Education & Reference · 12/12/2012
LHS : 2 csc 2u = 2 / (sin 2u) = 2 / (2 sin u cos u) = 1 / (sin u cos u) = (1/sin u) * (1/cos u) = sec u csc u = RHS
1 Answers · Education & Reference · 26/01/2014
LHS log (x-3) + log x can be written as log (x-3) multiplied with log x oh i just got confused.. need to help yew but.......... lol
4 Answers · Education & Reference · 13/04/2012
LHS = 5x - 5/8 RHS = 4x + 13/4 5x-5/8 = 4x + 13/4 5x - 4x = 13/4 + 5/8 x = 13/4 + 5/8 x = 26/8 + 5/8 x = 31/8
1 Answers · Education & Reference · 11/09/2013
1) LHS = tanx sinx + cosx = (sinx/cosx) sinx + cosx = (sin² x / cosx) + cos x = [(1...
1 Answers · Education & Reference · 27/04/2010
LHS : 4^(x-1)=(2^2)^(x-1)=2^(2x-2) RHS: 32^4=(2^5)^4=2^20 Therefore 20=2x-2 x=11
1 Answers · Education & Reference · 24/01/2014
LHS = sin x - 2 sin x cos x taking sin x common , we get sin x ( 1 - 2 cos x ) = RHS Good luck !
1 Answers · Education & Reference · 03/08/2012
By FOIL, it is 1 - sin^2 x which by the Pythagorean identity is cos^2 x
1 Answers · Education & Reference · 21/05/2013
tan²x • (csc²x – 1) = 1 sin²x • [ (1 ⁄ sin²x) – 1 ] ⁄ cos²x = 1 (1 – sin²x) ⁄ cos²x = 1 ... use identity: 1 – sin²x = cos²x...
1 Answers · Education & Reference · 21/05/2013
