If - and - are acute angles satisfying cos 2 -3 cos 2 -1-3-cos 2 - then tan-

The correct option is A √(2) tan β Given: cos 2 α =3 cos 2 β 1/3 cos 2 β ⇒ cos 2 α 1/cos 2 α + 1 =(3 cos 2β 1) (3 cos 2 β)/(3 cos 2 β 1)+(3 cos 2 β) (Usi ...

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Standard XII

Mathematics

Sin2A and Cos2A in Terms of tanA

If α and β ar...

Question

If $α a n d β$ are acute angles satisfying $c o s 2 α = \frac{3 c o s 2 β - 1}{3 - c o s 2 β},$ then $t a n α$ =

$\sqrt{2} t a n β$

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$\frac{1}{\sqrt{2}} t a n β$

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$\sqrt{2} c o t β$

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$\frac{1}{\sqrt{2}} c o t β$

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Solution

The correct option is A
$\sqrt{2} t a n β$

Given:

$c o s 2 α = \frac{3 c o s 2 β - 1}{3 - c o s 2 β} \Rightarrow \frac{c o s 2 α - 1}{c o s 2 α + 1} = \frac{(3 c o s 2 β - 1) - (3 - c o s 2 β)}{(3 c o s 2 β - 1) + (3 - c o s 2 β)}$

(Using componendo and dividendo)

$\Rightarrow \frac{c o s 2 α - 1}{c o s 2 α + 1} = \frac{4 c o s 2 β - 4}{2 c o s 2 β + 2} \Rightarrow - \frac{1 - c o s 2 α}{1 + c o s 2 α} = \frac{- 4 (1 - c o s 2 β)}{2 (1 + c o s 2 β)} \Rightarrow \frac{1 - c o s 2 α}{1 + c o s 2 α} = \frac{2 (1 - c o s 2 β)}{(1 + c o s 2 β)} \Rightarrow \frac{2 s i n^{2} α}{2 c o s^{2} α} = \frac{2 (2 s i n^{2} β)}{(2 c o s^{2} β)} \Rightarrow t a n^{2} α = 2 t a n^{2} β ∴ t a n α = \sqrt{2} t a n β$

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If α and β are acute angles satisfying cos 2α=3 cos 2β−13−cos 2β, then tan α =

If $α a n d β$ are acute angles satisfying $c o s 2 α = \frac{3 c o s 2 β - 1}{3 - c o s 2 β},$ then $t a n α$ =