Question

Prove
sinA/(secA+tanA-1) + cosA/(cosecA + cotA-1) = 1​

Answer 1

Step-by-step explanation:

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

• $\rm \dfrac{sinA}{(secA + tanA - 1)} + \dfrac{cosA}{(cosecA + cotA - 1)} = 1$

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

Let us prove by simplifying LHS.

$\rm LHS = \dfrac{sinA}{(secA + tanA - 1)} + \dfrac{cosA}{(cosecA + cotA - 1)}$

$\rm = \dfrac{sinA}{ \dfrac{1}{cosA} + \dfrac{sinA}{cosA} - 1} + \dfrac{cosA}{ \dfrac{1}{sinA }+ \dfrac{cosA}{sinA}- 1}$

$\rm = \dfrac{sinA}{ \dfrac{1 + sinA - cosA}{cosA} } + \dfrac{cosA}{ \dfrac{1 + cosA -sinA }{sinA }}$

$\rm = \dfrac{sinAcosA }{1 + sinA - cosA} + \dfrac{cosAsinA}{ {1 + cosA -sinA }}$

$\rm = sinAcosA \bigg[ \dfrac{1 }{1 + sinA - cosA} + \dfrac{1}{ {1 + cosA -sinA }} \bigg]$

$\rm = sinAcosA \bigg[ \dfrac{(1 + cosA -sinA) + (1 + sinA - cosA)}{(1 + sinA - cosA)(1 + cosA -sinA)} \bigg]$

$\rm = sinAcosA \bigg[ \dfrac{1 \: \: \cancel{+ cosA } \: \: \cancel{-sinA}+ 1 \: \: \cancel{+ sinA } \: \: \cancel{- cosA}}{1 +cosA - sinA + sinA cosA - {sin}^{2}A - cosA - cos^{2} A +sinA cosA} \bigg]$

$\rm = sinAcosA \bigg[ \dfrac {2}{1 - sin^{2} A - cos ^{2} A + 2sinAcosA} \bigg]$

$\rm = sinAcosA \bigg[ \dfrac {2}{1 -( sin^{2} A + cos ^{2} A )+ 2sinAcosA} \bigg]$

$\rm = sinAcosA \bigg[ \dfrac {2}{1 -1+ 2sinAcosA} \bigg] \: \: \: \: \: \: \: \: \: \: \: \bigg( \because {sin}^{2} A + cos ^{2} A = 1 \bigg)$

$\rm = \dfrac {2sinAcosA}{2sinAcosA}$

$\rm = 1 = RHS$

LHS = RHS

Hence Proved

$\bf\underline{\underline{\purple{Formulae\: used:-}}}$

• $\rm secA = \dfrac{1}{cosA}$

• $\rm tanA = \dfrac{ sinA}{cosA}$

• $\rm {sin}^{2} A + cos ^{2} A = 1$