Question

Prove that (secA/secA+1)+(secA/secA-1)=2cosec^2A​

Answer 1

Step-by-step explanation:

$\bf\underline{\underline{\red{To\:Prove:-}}}$

• $\rm \dfrac{secA}{secA + 1} + \dfrac{secA}{secA - 1} = 2 {cosec}^{2} A$

$\bf\underline{\underline{\green{Proof:-}}}$

Let us prove by simplifying LHS.

$\rm LHS = \dfrac{secA}{secA + 1} + \dfrac{secA}{secA - 1}$

$\rm = \dfrac{secA (secA - 1) + secA (secA + 1)}{(secA + 1)(secA - 1)}$

$\rm = \dfrac{sec^{2} A - secA + sec ^{2} A + secA }{(secA + 1)(secA - 1)}$

$\rm = \dfrac{sec^{2} A \: \: \cancel{- secA} + sec ^{2} A \: \: \cancel{+ secA} }{(secA)^{2} - {(1)}^{2} }$

$\rm = \dfrac{2sec^{2}A }{sec^{2}A - 1 }$

$\rm = \dfrac{2sec^{2}A }{tan^{2}A } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \because {sec}^{2} A - 1 = {tan}^{2} A \bigg)$

$\rm = \dfrac{2sec^{2}A }{ \: \: \: \: \: \dfrac{sin^{2}A}{cos^{2}A} \: \: \: \: } \: \: \: \: \: \: \: \: \: \: \bigg( \because tan^{2}A =\dfrac{sin^{2}A}{cos^{2}A} \bigg)$

$\rm = \dfrac{2sec^{2}A }{ sin^{2}A \: sec^{2}A } \: \: \: \: \: \: \: \: \: \: \bigg( \because \dfrac{1}{cos^{2}A} = sec^{2}A \bigg)$

$\rm = \dfrac{2 }{ \: sin^{2}A \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\bigg( \because \dfrac{1}{sin^{2}A} =cosec^{2}A \bigg)$

$\rm = 2 {cosec}^{2} A = RHS$

LHS = RHS

Hence Proved