[tex]\large {\underline{\sf{Solution}}} :[/tex]
[tex]\sf To \: prove : \sqrt{\dfrac{\sf secA - 1}{\sf secA + 1}} + \sqrt{\dfrac{\sf secA + 1}{\sf secA - 1}} = 2 \: cosecA[/tex]
we know that,
[tex]\: \: \: \: {\boxed{\sf (secA + 1) (secA - 1) = {tan}^{2} A}}[/tex]
From that we get
[tex]\sf secA + 1 = \dfrac{{tan}^{2} A}{secA - 1} \: \: \: (or) \: \: \: secA - 1 = \dfrac{{tan}^{2} A}{secA + 1}[/tex]
[tex]\rule{210pt}{2pt}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{\dfrac{\sf (secA - 1)^{2}}{\sf {tan}^{2} A}} + \sqrt{\dfrac{\sf(secA + 1)^{2}}{\sf {tan}^{2} A}} \\ = \sf \dfrac{secA - 1}{tanA} + \dfrac{secA + 1}{tanA} [/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \dfrac{2 \: secA}{sinA} cosA[/tex]
[tex]{\boxed{\sqrt{\dfrac{\sf secA - 1}{\sf secA + 1}} + \sqrt{\dfrac{\sf secA + 1}{\sf secA - 1}} = \sf 2 \: cosecA}}[/tex]
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[tex]\large {\underline{\sf{Solution}}} :[/tex]
[tex]\sf To \: prove : \sqrt{\dfrac{\sf secA - 1}{\sf secA + 1}} + \sqrt{\dfrac{\sf secA + 1}{\sf secA - 1}} = 2 \: cosecA[/tex]
we know that,
[tex]\: \: \: \: {\boxed{\sf (secA + 1) (secA - 1) = {tan}^{2} A}}[/tex]
From that we get
[tex]\sf secA + 1 = \dfrac{{tan}^{2} A}{secA - 1} \: \: \: (or) \: \: \: secA - 1 = \dfrac{{tan}^{2} A}{secA + 1}[/tex]
[tex]\rule{210pt}{2pt}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{\dfrac{\sf (secA - 1)^{2}}{\sf {tan}^{2} A}} + \sqrt{\dfrac{\sf(secA + 1)^{2}}{\sf {tan}^{2} A}} \\ = \sf \dfrac{secA - 1}{tanA} + \dfrac{secA + 1}{tanA} [/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \dfrac{2 \: secA}{sinA} cosA[/tex]
[tex]{\boxed{\sqrt{\dfrac{\sf secA - 1}{\sf secA + 1}} + \sqrt{\dfrac{\sf secA + 1}{\sf secA - 1}} = \sf 2 \: cosecA}}[/tex]