Question

please answer with explanation
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Answer 1

Answer:

hey here is your answer

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Step-by-step explanation:

$so \: here \: 3sin.theta = \sqrt{3} \: tan \: theta$

$ie \: 3sin.theta = \sqrt{3}.sin \: theta \div cos \: theta \\ as \: tan \: theta = sin \: theta \div cos \: theta \\ \\ ie \: 3 = \sqrt{3} \div cos \: theta \\ so \: cos \: theta = \sqrt{3} \div 3$

$now \: cos.square \: theta = ( \sqrt{3} \div 3)square \\ cos.square \: theta = 3 \div 9$

$so \: using \: \\ sin \: square \: theta = 1 - cos.square \: theta \\ = 1 - 3 \div 9 \\ = 9 - 3 \div 9 \\ = 6 \div 9$

$so \: thus \: then \\ sin.square \: theta - cos.square \: theta \\ = (6 \div 9) - (3 \div 9) \\ = 6 - 3 \div 9 \\ = 3 \div 9 \\ = 1 \div 3$

$hence \: final \: answer \: is \: 1 \div 3$