Question

prove that √ sec a -1 / sec a+1 + √ seca+1 /seca-1 = 2 cosec a​

Answer 1

Correct question :-

Given to prove that :-

$\sf \dfrac{\sqrt{secA-1} }{\sqrt{secA+1} } +\dfrac{\sqrt{secA+1} }{\sqrt{secA-1} }= 2cosecA$

Solution :-

Take L.H.S

$\sf \dfrac{\sqrt{secA-1} }{\sqrt{secA+1} } +\dfrac{\sqrt{secA+1} }{\sqrt{secA-1} }$

Take L.C.M to the denominator

$\sf \dfrac{(\sqrt{secA-1})(\sqrt{secA-1} )+(\sqrt{secA+1} )(\sqrt{secA+1} )}{(\sqrt{secA+1})(\sqrt{secA-1} ) }$

$\sf \dfrac{(\sqrt{secA+1} )^2+(\sqrt{secA-1})^2 }{(\sqrt{secA+1} )(\sqrt{secA-1}) }$

Applying (a+b)(a-b) formula to the denominator

(a+b)(a-b) = a²-b²

$\sf \dfrac{secA+1+secA-1}{\sqrt{(secA)^2-(1)^2} }$

$\sf \dfrac{2secA}{\sqrt{sec^2A-1} }$

We know that from Trigonometric identities ,

sec²A-tan²A = 1

sec²A -1 = tan²A

$\sf \dfrac{2secA}{\sqrt{tan^2A} }$

$\sf \dfrac{2secA}{tanA}$

From trigonometric relations ,

secA = 1/cosA

tanA = sinA/cosA

$\sf \dfrac{\dfrac{2}{cosA} }{\dfrac{sinA}{cosA} }$

$\sf \dfrac{2}{sinA}$

$\sf 2\bigg( \dfrac{1}{sinA} \bigg)$

From trigonometric relations ,

sinA = 1/cscA

$\sf = 2(cscA)$

$\sf = 2cscA$

Hence proved

Used formulae :-

Trigonometric relations :-

secA = 1/cosA

tanA = sinA/cosA

sinA = 1/cscA

Trigonometric identity :-

sec²A - tan²A = 1

sec²A- 1 = tan²A

Algebraic identity:-

(a+b)(a-b) = a²-b²

Answer 2

Correct question :-

Given to prove that :-

$\sf \dfrac{\sqrt{secA-1} }{\sqrt{secA+1} } +\dfrac{\sqrt{secA+1} }{\sqrt{secA-1} }= 2cosecA$

Solution :-

Take L.H.S

$\sf \dfrac{\sqrt{secA-1} }{\sqrt{secA+1} } +\dfrac{\sqrt{secA+1} }{\sqrt{secA-1} }$

Take L.C.M to the denominator

$\sf \dfrac{(\sqrt{secA-1})(\sqrt{secA-1} )+(\sqrt{secA+1} )(\sqrt{secA+1} )}{(\sqrt{secA+1})(\sqrt{secA-1} ) }$

$\sf \dfrac{(\sqrt{secA+1} )^2+(\sqrt{secA-1})^2 }{(\sqrt{secA+1} )(\sqrt{secA-1}) }$

Applying (a+b)(a-b) formula to the denominator

(a+b)(a-b) = a²-b²

$\sf \dfrac{secA+1+secA-1}{\sqrt{(secA)^2-(1)^2} }$

$\sf \dfrac{2secA}{\sqrt{sec^2A-1} }$

We know that from Trigonometric identities ,

sec²A-tan²A = 1

sec²A -1 = tan²A

$\sf \dfrac{2secA}{\sqrt{tan^2A} }$

$\sf \dfrac{2secA}{tanA}$

From trigonometric relations ,

secA = 1/cosA

tanA = sinA/cosA

$\sf \dfrac{\dfrac{2}{cosA} }{\dfrac{sinA}{cosA} }$

$\sf \dfrac{2}{sinA}$

$\sf 2\bigg( \dfrac{1}{sinA} \bigg)$

From trigonometric relations ,

sinA = 1/cscA

$\sf = 2(cscA)$

$\sf = 2cscA$

Hence proved

Used formulae :-

Trigonometric relations :-

secA = 1/cosA

tanA = sinA/cosA

sinA = 1/cscA

Trigonometric identity :-

sec²A - tan²A = 1

sec²A- 1 = tan²A

Algebraic identity:-

(a+b)(a-b) = a²-b²