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If 7sin²A+3cos²A=4, then prove that tanA=1/√3.

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Given

7sin^2A+3cos^2A=4

=>7sin^2A+3cos^2A=4(sin^2A+cos^2A)

=>3sin^2A=cos^2A

=>sin^2A/cos^2A=1/3

=>tan^2A=1/3

=>tanA=1/√3

Proved
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