Question

If tan^2alpha -2tan^2beta = 1.
Then which of the following is correct.

a) 2cos2beta - cos2apha = 1
b) cos2aplha - 2cos2beta = 1
c) 2cos2apha - cos2beta = -1
d) cos2beta - 2cos2apha = -1

Answer 1

i think c is the answer of ur question

Answer 2

Hello Dear,

Thanks For Asking the Question :)

Your Answer is Below

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Question :-

$\tan ^2\left(\alpha \right)-2\tan ^2\left(\beta \right)\:=\:1$

Explanation :-

• shift $-2\tan ^2\left(\beta \right)\:$ to RHS

⇒ $\tan ^2\left(\alpha \right)\:=\:1+2\tan ^2\left(\beta \right)\:$

• add 1 on both sides

⇒ $1+\tan ^2\left(\alpha \right)\:=\:1+2\tan ^2\left(\beta \right)\:+1$

⇒ $\sec ^2\left(\alpha\right)-2\sec ^2\left(\beta\right)=0$

⇒ $\frac{1}{cos^{2}(\alpha) } = \frac{2}{cos^{2}(\beta) }$

• cross multiply

⇒ $cos^{2} (\beta) = 2\ cos^{2} (\alpha)$

• $2cos^{2} \theta - 1 = cos2\theta \ \ \ ->\ \ \ 2cos^{2} \theta=cos2\theta+1$

• $\huge{cos^{2} \theta = \huge{ \frac{1+cos2\theta}{2} }}$

⇒ $\frac{1+cos2\beta}{2} = 1+cos2\alpha$

• cross multiply

⇒ $1\ +\ cos2\beta = 2+2cos\ 2\alpha$

⇒ $cos2\beta= 2cos\ 2\alpha +2-1$

⇒ $cos2\beta - 2cos\ 2\alpha = 1$

• multiply - on both sides

⇒ $[-(cos2\beta) ]\ [-(- 2cos\ 2\alpha )] = -(1)$

⇒ $-cos2\beta + 2cos\ 2\alpha = -1$

⇒ $2cos\ 2\alpha - cos2\beta = -1$

Option c) $2cos\ 2\alpha - cos2\beta = -1$  is the correct answer

      $\huge{\orange{\boxed{\boxed{\pink{\underline{\green{\mathscr{Option\ c)\ 2cos2\alpha - cos2\beta = -1 }}} }} }}}$

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Hope It Helps You Dear ! :D     ^_^

By Abhiram5566

                                  MARK ME AS BRAINLIEST ANSWER

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