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If then prove that tanA = 1/√3​

Answer»

To PROVE :-

  • \sf{tanA=\frac{1}{\sqrt{3}}}

PROOF :-

GIVEN,

\sf{7sin^2A+3cos^2A=4}

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\sf{7sin^2A+3cos^2A=4}

:\implies\sf{7sin^2A+3(1-sin^2A)=4}

:\implies\sf{7sin^2A+3-3sin^2A=4}

:\implies\sf{4sin^2A=1}

:\implies\sf{sin^2A=\frac{1}{4}}

:\implies\sf{sinA=\sqrt{\frac{1}{4}}=\frac{1}{2}}

:\implies\sf{sinA=sin30^{\circ}}

:\implies\sf{A=30^{\circ}}

\therefore\sf{tanA=tan30^{\circ}=\frac{1}{\sqrt{3}}}

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