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7sin^2A+3cos^2A=4 show that tanA=1/root3

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Posted by Priya Patel 7 years, 10 months ago

Naveen Sharma 7 years, 10 months ago

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

$=> 4sin^2A+ 3sin^2A+3cos^2A=4$

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

$=> 4sin^2A+ 3=4$
$=> 4sin^2A = 1$

$=> sin^2A = {1\over 4}$

$=> sinA = {1\over 2}$

$=> sinA = sin 30^o$

$=> A = 30^o$

So $tan A= tan 30^o = {1\over \sqrt 3}$

ANSWER

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