At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
[tex]\large\underline{\sf{Solution-}} [/tex]
Given expression is
[tex]\rm :\longmapsto\:\prod \limits^{ \infty }_{n = 1} \bigg( \dfrac{25}{777} \bigg)^{ {( - 1)}^{n} \bigg( \dfrac{ tan^{4n} (x) + 1}{ tan ^{2n} (x) } \bigg)}[/tex]
can be rewritten as
[tex]\rm \: = \: \prod \limits^{ \infty }_{n = 1} \bigg( \dfrac{25}{777} \bigg)^{ {( - 1)}^{n} \bigg( \dfrac{ tan^{4n} (x)}{ tan ^{2n} (x)} + \dfrac{1}{ {tan}^{2n} (x)} \bigg)}[/tex]
[tex]\rm \: = \: \prod \limits^{ \infty }_{n = 1} \bigg( \dfrac{25}{777} \bigg)^{ {( - 1)}^{n} \bigg( {tan}^{2n}(x) + {cot}^{2n}(x) \bigg)}[/tex]
can be further rewritten as
[tex]\begin{gathered}\rm \: = \: \prod \limits^{ \infty }_{n = 1} \bigg( \dfrac{25}{777} \bigg)^{ {( - 1)}^{n} \bigg( {( {tan}^{2}x) }^{n} + {( {cot}^{2} x)}^{n} \bigg)} \\ \end{gathered} [/tex]
[tex]\begin{gathered}\rm \: = \: \prod \limits^{ \infty }_{n = 1} \bigg( \dfrac{25}{777} \bigg)^{\bigg( { {( - 1)}^{n} ( {tan}^{2}x) }^{n} + {( - 1)}^{n} {( {cot}^{2} x)}^{n} \bigg)} \\ \end{gathered}[/tex]
[tex]\begin{gathered}\rm \: = \: \prod \limits^{ \infty }_{n = 1} \bigg( \dfrac{25}{777} \bigg)^{ \bigg( {( { - tan}^{2}x) }^{n} + {( { - cot}^{2} x)}^{n} \bigg)} \\ \end{gathered} [/tex]
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{ \bigg( {( { - tan}^{2}x) }^{1} + {( { - cot}^{2} x)}^{1} \bigg)} \times \bigg( \dfrac{25}{777} \bigg)^{ \bigg( {( { - tan}^{2}x) }^{2} + {( { - cot}^{2} x)}^{2} \bigg)} \times \bigg( \dfrac{25}{777} \bigg)^{ \bigg( {( { - tan}^{2}x) }^{3} + {( { - cot}^{2} x)}^{3} \bigg)} - - - \\ \end{gathered} [/tex]
[tex]\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{ \bigg( - {tan}^{2}x - {cot}^{2}x + {tan}^{4}x + {cot}^{4}x - {tan}^{6}x - {cot}^{6}x + - - \bigg)}[/tex]
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{ \bigg(( - {tan}^{2}x + {tan}^{4}x - {tan}^{6}x + - - ) + ( - {cot}^{2}x + {cot}^{4}x - {cot}^{6}x + - -) \bigg)} \\ \end{gathered} [/tex]
Using Sum of infinite GP series,
[tex]\begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ S_ \infty \: = \: \frac{a}{1 - r} \: where \: |r| < 1}}} \\ \end{gathered}[/tex]
So, using this,
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{ \bigg(\dfrac{ - {tan}^{2}x}{1 + {tan}^{2}x} + \dfrac{ - {cot}^{2}x}{1 + {cot}^{2}x} \bigg)} \\ \end{gathered} [/tex]
We know,
[tex]\begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ 1 + {tan}^{2}x = {sec}^{2}x}}} \\ \end{gathered} [/tex]
and
[tex]\begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ 1 + {cot}^{2}x = {cosec}^{2}x}}} \\ \end{gathered} [/tex]
So, using this, we get
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{ \bigg(\dfrac{ - {tan}^{2}x}{{sec}^{2}x} + \dfrac{ - {cot}^{2}x}{{cosec}^{2}x} \bigg)} \\ \end{gathered} [/tex]
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{ \bigg(\dfrac{ - {sin}^{2}x}{{cos}^{2}x} \times {cos}^{2}x + \dfrac{ - {cos}^{2}x}{{sin}^{2}x} \times {sin}^{2}x \bigg)} \\ \end{gathered} [/tex]
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{\bigg( - {sin}^{2}x - {cos}^{2}x \bigg)} \\ \end{gathered} [/tex]
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{\bigg( -({sin}^{2}x + {cos}^{2}x) \bigg)} \\ \end{gathered} [/tex]
[tex]\begin{gathered}\rm \: = \: \bigg( \dfrac{25}{777} \bigg)^{\bigg( -1 \bigg)} \\ \end{gathered} [/tex]
[tex]\rm \: = \: \dfrac{777}{25}[/tex]
Hence,
[tex]\begin{gathered} \\ \purple{\rm :\longmapsto\:\boxed{\tt{ \prod \limits^{ \infty }_{n = 1} \bigg( \frac{25}{777} \bigg)^{ {( - 1)}^{n} \bigg( \frac{ tan^{4n} (x) + 1}{ tan ^{2n} (x) } \bigg)} = \frac{777}{25} \: }}} \\ \end{gathered} [/tex]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.