1-sinx/1+sinx=(secx-tanx)²
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Si loo caddeeyo aqoonsigan, waxaanu ku bilaabi doonaa dhinaca bidixda (LHS) ee isla’egta waxaanan ku maamuli doonaa si aanu u helno dhinaca midigta (RHS):
LHS: (1-sinx)/(1+ sinx)
Talaabada 1: Ku dhufo tirada iyo hooseeyaha (1-sinx):
LHS: (1-sinx)/(1+sinx) * (1-sinx)/(1-sinx)
LHS: (1 – 2sinx + sin²x)/(1 – sin²x)
Tallaabada 2: Isticmaal aqoonsiga sin²x + cos²x = 1 si aad ugu badasho cos²x 1 – sin²x:
LHS: (1 – 2sinx + sin²x)/cos²x
Talaabada 3: Isticmaal aqoonsiga sexx = 1/cosx iyo tanx = six/cosx si aad ugu qorto RHS xaga sexx iyo tanx:
RHS: (sx – tanx)²
RHS: (1/cosx – six/cosx)²
RHS: ((1-sinx)/cosx)²
RHS: (1 – 2sinx + sin²x)/cos²x
Hadda waxaan haynaa tibaax isku mid ah labada LHS iyo RHS. Sidaa darteed, waxaanu caddaynay in:
(1-sinx)/(1+sinx) = (sx-tanx)²