If 7 sin2theta + 3 cos2 theta=4 show that tan theta = 1/root3
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Answered by
54
7sin2ϴ + 3cos2ϴ = 4
4sin2 ϴ + 3sin2 ϴ+ 3cos2 ϴ = 4
4sin2 ϴ +3 (sin2 ϴ + cos2 ϴ) = 4
4sin2 ϴ + 3 = 4
4sin2 ϴ = 1
sin2 ϴ =1/4
sin ϴ = ½
sin ϴ =perp./hypo. = ½
So, perp. = 1k ; hypo. = 2k
(Hypo)2 = (base)2 + (perp.)2
k2 = (base)2 + (2k)2
(base)2 = 4k2 – k2
base = root3 k
tan ϴ = perp./base = k/root3 k
so, tan ϴ =1/root3
Hope it helps !!!!!!!!!!!!!!!!!!!!!!!!!
4sin2 ϴ + 3sin2 ϴ+ 3cos2 ϴ = 4
4sin2 ϴ +3 (sin2 ϴ + cos2 ϴ) = 4
4sin2 ϴ + 3 = 4
4sin2 ϴ = 1
sin2 ϴ =1/4
sin ϴ = ½
sin ϴ =perp./hypo. = ½
So, perp. = 1k ; hypo. = 2k
(Hypo)2 = (base)2 + (perp.)2
k2 = (base)2 + (2k)2
(base)2 = 4k2 – k2
base = root3 k
tan ϴ = perp./base = k/root3 k
so, tan ϴ =1/root3
Hope it helps !!!!!!!!!!!!!!!!!!!!!!!!!
Answered by
38
7sin²θ + 3cos²θ = 4
7sin²θ + ౩(1-sin²θ) = 4
7sin²θ+3-3sin²θ = 4
4sin²θ+3 = 4
4sin²θ=1
sin²θ = 1/4
sinθ = +1/2
sinθ = + π/6
Therefore tan θ = + π/6
then we get tan θ = + 1/√3
7sin²θ + ౩(1-sin²θ) = 4
7sin²θ+3-3sin²θ = 4
4sin²θ+3 = 4
4sin²θ=1
sin²θ = 1/4
sinθ = +1/2
sinθ = + π/6
Therefore tan θ = + π/6
then we get tan θ = + 1/√3
akhilsatish:
so?
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