Question

cos 2 alpha is equal to 3 cos square beta - 1 / 3 - cos 2 beta therefore find tan alpha

Answer 1

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Answer 2

Answer:

Value of $\frac{tan\,\alpha}{tan\,\beta}\:\:is\:\:\sqrt{2}$

Step-by-step explanation:

Given: $cos\,2\aplha=\frac{3cos^2\,\beta-1}{3-cos\,2\beta}$

To find: $\frac{tan\,\alpha}{tan\,\beta}$

Consider,

$cos\,2\aplha=\frac{3cos^2\,\beta-1}{3-cos\,2\beta}$

$\frac{1-cos\,2\alpha}{1+cos\,2\alpha}=\frac{3-cos\,2\beta-3cos\,2\beta+1}{3-cos\,2\beta+3cos\,2\beta-1}$

$tan^2\,\alpha=\frac{4-4cos\,2\beta}{2+2cos\,2\beta}$

$tan^2\,\alpha=2\times\frac{2-2cos\,2\beta}{1+cos\,2\beta}$

$tan^2\,\alpha=2\times tan^2\,\beta}$

$\frac{tan^2\,\alpha}{tan^2\,\beta}=2$

$(\frac{tan\,\alpha}{tan\,\beta})^2=2$

$\frac{tan\,\alpha}{tan\,\beta}=\sqrt{2}$

$\frac{tan\,\alpha}{tan\,\beta}=\frac{\sqrt{2}}{1}$

Therefore, Value of $\frac{tan\,\alpha}{tan\,\beta}\:\:is\:\:\sqrt{2}$