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## QuestionMathsClass 10

If $$\sqrt{3}tan\theta =3sin\theta ,$$ then the value of $$si{n}^{2}\theta -co{s}^{2}\theta$$ is
(A) $$\frac{1}{2}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{1}{4}$$
(D) $$\frac{1}{5}$$

(B)
4.6
4.6

## Solution

$$\sqrt{3}tan\theta =3sin\theta$$
$$\sqrt{3 }\frac{sin\theta }{cos\theta }=3sin\theta$$
$$cos\theta =\frac{\sqrt{3}}{3}=cos\theta =\frac{1}{\sqrt{3}}$$
$$\therefore\;si{n}^{2}\theta +co{s}^{2}\theta =1$$
$$si{n}^{2}\theta =1-co{s}^{2}\theta$$
$$=si{n}^{2}\theta -co{s}^{2}\theta$$
$$=1-co{s}^{2}\theta -co{s}^{2}\theta$$
$$=1-2\times {\left(\frac{1}{\sqrt{3}}\right)}^{2}$$
$$=1-2\times \frac{1}{3}=1-\frac{2}{3}=\frac{1}{3}$$