Math, asked by SharmaShivam, 10 months ago

If \sf{tan\:\theta\:tan\:\varnothing\:=\:\sqrt{\dfrac{a-b}{a+b}}}, prove that
\sf{\left(a\:-\:b\:cos\:2\theta\right)\left(a\:-\:b\:cos\:2\varnothing\right)} is independent of \theta and \varnothing.​

Answers

Answered by shadowsabers03
3

Given,

\displaystyle\longrightarrow\sf{\tan\theta\tan\phi=\sqrt{\dfrac{a-b}{a+b}}}

\displaystyle\longrightarrow\sf{\tan^2\theta\tan^2\phi=\dfrac{a-b}{a+b}\quad\quad\dots(1)}

\displaystyle\longrightarrow\sf{a-b=(a+b)\tan^2\theta\tan^2\phi\quad\quad\dots(2)}

We see that,

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=a-b\cdot\dfrac{1-\tan^2\theta}{1+\tan^2\theta}}

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=a-\dfrac{b-b\tan^2\theta}{1+\tan^2\theta}}

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=\dfrac{a+a\tan^2\theta-b+b\tan^2\theta}{1+\tan^2\theta}}

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=\dfrac{a-b+a\tan^2\theta+b\tan^2\theta}{1+\tan^2\theta}}

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=\dfrac{(a-b)+(a+b)\tan^2\theta}{1+\tan^2\theta}}

Substituting value of \sf{a-b} from (2),

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=\dfrac{(a+b)\tan^2\theta\tan^2\phi+(a+b)\tan^2\theta}{1+\tan^2\theta}}

\displaystyle\longrightarrow\sf{a-b\cos(2\theta)=\dfrac{(1+\tan^2\phi)(a+b)\tan^2\theta}{1+\tan^2\theta}}

Similarly,

\displaystyle\longrightarrow\sf{a-b\cos(2\phi)=\dfrac{(1+\tan^2\theta)(a+b)\tan^2\phi}{1+\tan^2\phi}}

Then,

\displaystyle\longrightarrow\sf{(a-b\cos(2\theta))(a-b\cos(2\phi))=\dfrac{(1+\tan^2\phi)(a+b)\tan^2\theta}{1+\tan^2\theta}\cdot\dfrac{(1+\tan^2\theta)(a+b)\tan^2\phi}{1+\tan^2\phi}}

\displaystyle\longrightarrow\sf{(a-b\cos(2\theta))(a-b\cos(2\phi))=(a+b)^2\tan^2\theta\tan^2\phi}

From (1),

\displaystyle\longrightarrow\sf{(a-b\cos(2\theta))(a-b\cos(2\phi))=(a+b)^2\cdot\dfrac{a-b}{a+b}}

\displaystyle\longrightarrow\sf{(a-b\cos(2\theta))(a-b\cos(2\phi))=(a+b)(a-b)}

\displaystyle\longrightarrow\sf{\underline{\underline{(a-b\cos(2\theta))(a-b\cos(2\phi))=a^2-b^2}}}

Hence we see that \sf{(a-b\cos(2\theta))(a-b\cos(2\phi))} is independent of \theta and \phi.

Hence Proved!

Answered by wwwseenalingampalli
0

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