Math, asked by sajan6491, 20 hours ago

  \large \displaystyle\sf \prod^{ \infty }_{n = 1} \bigg( \frac{25}{777}  \bigg)^{ {( - 1)}^{n}  \bigg( \frac{ \tan^{4n} (x) + 1}{ \tan ^{2n} (x)  } \bigg)  }

Answers

Answered by senboni123456
15

Step-by-step explanation:

We have,

\tt{\displaystyle\,P=\prod^{\infty}_{n=1}\,\left(\dfrac{25}{777}\right)^{\displaystyle(-1)^n\left(\dfrac{tan^{4n}(x)+1}{tan^{2n}(x)}\right)}}

\tt{\displaystyle\,\implies\,P=\prod^{\infty}_{n=1}\,\left(\dfrac{25}{777}\right)^{\displaystyle(-1)^n\left(tan^{2n}(x)+cot^{2n}(x)\right)}}

\tt{\displaystyle\,\implies\,P=\prod^{\infty}_{n=1}\,\left(\dfrac{25}{777}\right)^{\displaystyle(-1)^n\left\{\left(tan^{2}(x)\right)^n+\left(cot^{2}(x)\right)^n\right\}}}

\tt{\displaystyle\,\implies\,P=\prod^{\infty}_{n=1}\,\left(\dfrac{25}{777}\right)^{\displaystyle\left\{(-1)^n\cdot\left(tan^{2}(x)\right)^n+(-1)^n\cdot\left(cot^{2}(x)\right)^n\right\}}}

\tt{\displaystyle\,\implies\,P=\prod^{\infty}_{n=1}\,\left(\dfrac{25}{777}\right)^{\displaystyle\left\{\left(-tan^{2}(x)\right)^n+\left(-cot^{2}(x)\right)^n\right\}}}

Now, consider the proposition,

\bigstar\,\,\sf{\displaystyle\prod^{\infty}_{m=1}\big(k\big)^{\displaystyle\,a^{n}}}

\mapsto\,\,\sf{\displaystyle\big(k\big)^{\displaystyle\,a}\cdot\big(k\big)^{\displaystyle\,a^2}\cdot\big(k\big)^{\displaystyle\,a^3}\cdot\big(k\big)^{\displaystyle\,a^4}\cdots}

\mapsto\,\,\sf{\displaystyle\big(k\big)^{\displaystyle\,a+a^2+a^3+a^4+\cdots}}

\mapsto\,\,\sf{\displaystyle\big(k\big)^{\displaystyle\,\sum^{\infty}_{m=1}\,a^m}}

So, our series may be rewritten as,

\tt{\displaystyle\,\implies\,P=\left(\dfrac{25}{777}\right)^{\displaystyle\sum^{\infty}_{n=1}\left\{\left(-tan^{2}(x)\right)^n+\left(-cot^{2}(x)\right)^n\right\}}}

Let us now consider the sum,

\tt{\displaystyle\,S=\sum^{\infty}_{n=1}\left\{\left(-tan^{2}(x)\right)^n+\left(-cot^{2}(x)\right)^n\right\}}

\tt{\displaystyle\,\implies\,S=\sum^{\infty}_{n=1}\left(-tan^{2}(x)\right)^n+\sum^{\infty}_{n=1}\left(-cot^{2}(x)\right)^n}

\bf{Put\,\,\,\, \alpha=-tan^2(x)\,\,\,\,\,\,and\,\,\,\,\,\,\beta=-cot^2(x)}

So,

\tt{\displaystyle\,\implies\,S=\sum^{\infty}_{n=1}\left(\alpha\right)^n+\sum^{\infty}_{n=1}\left(\beta\right)^n}

\tt{\displaystyle\,\implies\,S=\sum^{\infty}_{n=1}\left(\alpha\right)^n+\sum^{\infty}_{n=1}\left(\beta\right)^n}

These are GPs

So,

\tt{\displaystyle\,\implies\,S=\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}}

Putting the values of \alpha and \beta

\tt{\displaystyle\,\implies\,S=\dfrac{-tan^2(x)}{1+tan^2(x)}+\dfrac{-cot^2(x)}{1+cot^2(x)}}

\tt{\displaystyle\,\implies\,S=-\left(\dfrac{tan^2(x)}{sec^2(x)}+\dfrac{cot^2(x)}{cosec^2(x)}\right)}

\tt{\displaystyle\,\implies\,S=-\left(sin^2(x)+cos^2(x)\right)}

\tt{\displaystyle\,\implies\,S=-\left(1\right)}

\tt{\displaystyle\,\implies\,S=-1}

\tt{\displaystyle\,\therefore\sum^{\infty}_{n=1}\left\{\left(-tan^{2}(x)\right)^n+\left(-cot^{2}(x)\right)^n\right\}=-1}

So, our series becomes,

\tt{\displaystyle\,\implies\,P=\left(\dfrac{25}{777}\right)^{\displaystyle\sum^{\infty}_{n=1}\left\{\left(-tan^{2}(x)\right)^n+\left(-cot^{2}(x)\right)^n\right\}}=\left(\dfrac{25}{777}\right)^{-1}}

\tt{\displaystyle\,\implies\,P=\dfrac{777}{25}}

Hope this will help you!! :):)

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