Question

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If 3 tan²θ-1=0,cos 2θ will be??

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Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{3\,tan^2(\theta) - 1 = 0}$

$\tt{\implies\,tan^2(\theta)=\dfrac{1}{3}}$

Now,

$\tt{cos(2\,\theta)=\dfrac{1-tan^2(\theta)}{1+tan^2(\theta)}}$

$\tt{\implies\,cos(2\,\theta)=\dfrac{1-\dfrac{1}{3}}{1+\dfrac{1}{3}}}$

$\tt{\implies\,cos(2\,\theta)=\dfrac{\dfrac{3-1}{3}}{\dfrac{3+1}{3}}}$

$\tt{\implies\,cos(2\,\theta)=\dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}}$

$\tt{\implies\,cos(2\,\theta)=\dfrac{2}{4}}$

$\tt{\implies\,cos(2\,\theta)=\dfrac{1}{2}}$