If √3tan theta=3sin theta,prove that sin^2 theta-cos^2=1/3
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let theta=x.......
given that,
√3tanx=3sinx
√3tanx=(3sinx/cosx)cosx
√3tanx=3tanx.cosx
√3=3cosx
cosx=√3/3
(cosx)^2=(√3/3)^2
(cosx)^2=1/3
And ,
1-(sinx)^2=1/3 [(cosx)^2=1-(sinx)^2]
(sinx)^2=1-1/3=2/3
so,
(sinx)^2-(cosx)^2=2/3-1/3
(sinx)^2-(cosx)^2=1/3 [proved]
given that,
√3tanx=3sinx
√3tanx=(3sinx/cosx)cosx
√3tanx=3tanx.cosx
√3=3cosx
cosx=√3/3
(cosx)^2=(√3/3)^2
(cosx)^2=1/3
And ,
1-(sinx)^2=1/3 [(cosx)^2=1-(sinx)^2]
(sinx)^2=1-1/3=2/3
so,
(sinx)^2-(cosx)^2=2/3-1/3
(sinx)^2-(cosx)^2=1/3 [proved]
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