If 1+sin^2 theta = 3sin theta cos theta, then prove that tan theta =1or 1/2
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1 + sin^2 Ø = 3sinØcosØ
divide by cosØ
1/cosØ + tanØsinØ = 3sinØ
secØ + tanØsinØ = 3sinØ
tanØ = (3sinØ - secØ)/sinØ
= 3 - secØ/sinØ
let's stop here:
tanØ = 3 - 1/(sinØcosØ) ******
if tanØ = 1 , sinØ/cosØ = 1 and sinØ = cosØ, Ø = π/2
in *****
LS = tanØ = 1
RS = 3 - 1/sin^2 Ø
= 3 - 1/(1/2) = 3-2 = 1
= LS
if tanØ = 1/2
sinØ = 1/√5 and cosØ = 2/√5
in *****
LS = 1/2
RS = 3 - 1/(sinØcosØ)
= 3 - 1/((1/√5)(2/√5))
= 3 - 1/(2/5)
= 3 - 5/2
= 1/2
= LS
[Note:-Ø represents theta]
divide by cosØ
1/cosØ + tanØsinØ = 3sinØ
secØ + tanØsinØ = 3sinØ
tanØ = (3sinØ - secØ)/sinØ
= 3 - secØ/sinØ
let's stop here:
tanØ = 3 - 1/(sinØcosØ) ******
if tanØ = 1 , sinØ/cosØ = 1 and sinØ = cosØ, Ø = π/2
in *****
LS = tanØ = 1
RS = 3 - 1/sin^2 Ø
= 3 - 1/(1/2) = 3-2 = 1
= LS
if tanØ = 1/2
sinØ = 1/√5 and cosØ = 2/√5
in *****
LS = 1/2
RS = 3 - 1/(sinØcosØ)
= 3 - 1/((1/√5)(2/√5))
= 3 - 1/(2/5)
= 3 - 5/2
= 1/2
= LS
[Note:-Ø represents theta]
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