Question

3 - 4 sin squared theta by cos square theta

Answer 1

Find your answer in the attachment

Answer 2

$\frac{3 - 4sin^{2} \theta }{cos^{2} \theta }$

$Dividing\: numerator \:and \:denominator\\ by \: \blue { cos^{2} \theta} , we \:get$

$= \frac{\frac{(3 - 4sin^{2} \theta)}{cos^{2} \theta }}{\frac{cos^{2} \theta }{cos^{2} \theta}}$

$= \frac{3}{cos^{2} \theta} - \frac{4sin^{2} \theta }{cos^{2} \theta } \\= 3sec^{2} \theta - 4tan^{2} \theta \\= 3(1+tan^{2} \theta ) - 4tan^{2} \theta \\= 3 + 3tan^{2} \theta - 4tan^{2} \theta \\= 3 - tan^{2} \theta$

Therefore.,

$\red{\frac{3 - 4sin^{2} \theta }{cos^{2} \theta }}\\\green {= 3 - tan^{2} \theta}$

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