Question

If $\bold{7sin^2A+3cos^2A=4}$
then prove that tanA = 1/√3​

Answer 1

To Prove :-

• $\sf{tanA=\frac{1}{\sqrt{3}}}$

Proof :-

Given,

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

_________

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

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

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

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

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

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

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

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

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

• Hence,Proved

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