[tex] \large\underline{\sf{Solution-}}[/tex]

Consider RHS

[tex]\sf \: (1 + sec2\phi)(1 + sec {2}^{2}\phi)(1 + sec {2}^{3}\phi)...(1 + sec {2}^{n}\phi ) \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \bigg(1 + \dfrac{1}{cos2\phi } \bigg)\bigg(1 + \dfrac{1}{cos {2}^{2} \phi } \bigg)\bigg(1 + \dfrac{1}{cos {2}^{3} \phi } \bigg)...\bigg(1 + \dfrac{1}{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \bigg(\dfrac{cos2\phi + 1}{cos2\phi } \bigg)\bigg(\dfrac{cos {2}^{2} \phi + 1}{cos {2}^{2} \phi } \bigg)\bigg(\dfrac{cos {2}^{3} \phi + 1}{cos {2}^{3} \phi } \bigg)...\bigg(\dfrac{cos {2}^{n} \phi + 1}{cos {2}^{n} \phi } \bigg)\\ \\ [/tex]

[tex]\sf \: = \bigg(\dfrac{2 {cos}^{2}\phi }{cos2\phi } \bigg)\bigg(\dfrac{2 {cos}^{2}2\phi }{cos {2}^{2} \phi } \bigg)\bigg(\dfrac{2 {cos}^{2} {2}^{2} \phi }{cos {2}^{3} \phi } \bigg)...\bigg(\dfrac{2 {cos}^{2} {2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: cos2x = {2cos}^{2}x - 1 \: \: }} \\ \\ [/tex]

[tex]\sf \: = 2 {cos}^{2}\phi (2 {cos}^{}2\phi )(2 {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

can be further rewritten as

[tex]\sf \: = cos\phi (2 {cos}^{}\phi) (2 {cos}^{}2\phi )(2 {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{cos\phi}{sin\phi } (2sin\phi {cos}^{}\phi) (2 {cos}^{}2\phi )(2 {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{tan\phi } (sin2\phi ) (2 {cos}^{}2\phi )(2 {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: 2 \: sinx \: cosx \: = \: sin2x \: }} \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{tan\phi } (2sin2\phi {cos}^{}2\phi)(2 {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{tan\phi } (sin2^{2}\phi)(2 {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{tan\phi } (2sin2^{2}\phi {cos}2^{2}\phi ) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{tan\phi } (sin2^{3}\phi) ...\bigg(\dfrac{2 {cos}{2}^{n - 1} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{tan\phi } \bigg(\dfrac{{sin}{2}^{n} \phi }{cos {2}^{n} \phi } \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{tan {2}^{n} \phi }{tan\phi } \\ \\ [/tex]

Hence,

[tex]\boxed{ \sf{ \:(1 + sec2\phi)(1 + sec {2}^{2}\phi)(1 + sec {2}^{3}\phi)...(1 + sec {2}^{n}\phi ) = \dfrac{tan {2}^{n}\phi }{tan\phi } \: }} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\qquad\begin{gathered} { \boxed{ \begin{array}{c} & \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered} \\ \\ [/tex]

( tan 2ⁿ Φ )/tan Φ = (1 + sec 2Φ)(1 + sec 2²Φ)(1 + sec 2³Φ)(1 + sec 2⁴Φ)......... (1 + sec 2ⁿΦ)

Hello

Divya

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Missing U in brainly

U r not coming online

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