Verified answer

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

Consider,

[tex]\sf \: \sqrt {\dfrac{secA - 1}{secA + 1} }+ \sqrt{ \dfrac{secA + 1}{secA - 1}} \\ [/tex]

[tex]\sf \: = \: \dfrac{ \sqrt{secA - 1} }{ \sqrt{secA + 1} } + \dfrac{ \sqrt{secA + 1} }{ \sqrt{secA - 1} } \\ [/tex]

[tex]\sf \: = \: \dfrac{secA - 1 + secA + 1}{ \sqrt{secA + 1} \: \sqrt{secA - 1} } \\ [/tex]

[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{(secA + 1)(secA + 1)} \: } \\ [/tex]

[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {sec}^{2}A - {1}^{2} } } \\ [/tex]

[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {sec}^{2}A - 1} } \\ [/tex]

[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {tan}^{2}A } } \\ [/tex]

[tex]\sf \: = \: \dfrac{2secA}{ {tan}A } \\ [/tex]

[tex]\sf \: = \: \dfrac{2 \times \dfrac{1}{cosA} }{\dfrac{sinA}{cosA} } \\ [/tex]

[tex]\sf \: = \: \dfrac{2 }{sinA } \\ [/tex]

[tex]\sf \: = \: 2 \: cosecA[/tex]

Hence,

[tex]\implies\sf \: \boxed{\bf \: \sqrt {\dfrac{secA - 1}{secA + 1}} + \sqrt {\dfrac{secA + 1}{secA - 1} } = \: 2 \: cosecA \: } \\ [/tex]

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

Formulae Used:

[tex]\sf \: \sqrt{x} \times \sqrt{y} = \sqrt{xy} \\ [/tex]

[tex]\sf \: (x + y)(x - y) = {x}^{2} - {y}^{2} \\ [/tex]

[tex]\sf \: {sec}^{2}A - {tan}^{2}A = 1 \\ [/tex]

[tex]\sf \: secA = \dfrac{1}{cosA} \\ [/tex]

[tex]\sf \: tanA = \dfrac{sinA}{cosA} \\ [/tex]

[tex]\sf \: cosecA = \dfrac{1}{sinA} \\ [/tex]