Given that $$ an \alpha \cdotp an \beta = \dfrac{\sqrt{3} }{3}$$, find the value of $$ (2-\cos 2\alpha )(2-\cos 2\beta) $$

Question

Given that $$	an \alpha \cdotp 	an \beta = \dfrac{\sqrt{3}  }{3}$$, find the value of $$ (2-\cos 2\alpha )(2-\cos 2\beta)   $$

Uniteasy Admin · Accepted Answer

$$\cos 2\alpha = \dfrac{1-	an^2 \alpha }{1+	an^2 \alpha}  $$
$$2-\cos 2\alpha = 2-\dfrac{1-	an^2 \alpha }{1+	an^2 \alpha}$$
$$=\dfrac{2+2	an^2 \alpha-1+	an^2 \alpha}{1+	an^2 \alpha} $$
$$=\dfrac{1+3 	an^2 \alpha}{1+	an^2 \alpha} $$
Similarly, 
$$2-\cos 2\beta = \dfrac{1+3 	an^2 \beta }{1+	an^2 \beta } $$
$$ (2-\cos 2\alpha )(2-\cos 2\beta)  $$ 
$$=\dfrac{1+3 	an^2 \alpha}{1+	an^2 \alpha}\cdotp  \dfrac{1+3 	an^2 \beta }{1+	an^2 \beta } $$
$$=\dfrac{1+9	an^2 \alpha	an^2 \beta+3	an^2 \alpha+3	an^2 \beta}{1+	an^2 \alpha	an^2 \beta+	an^2 \alpha+	an^2 \beta} $$
$$=\dfrac{4+3	an^2 \alpha+3	an^2 \beta}{\dfrac{4}{3} +	an^2 \alpha+	an^2 \beta } =3$$

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