Question

Prove that : ( sinA+secA)^2 +(cosA+cosecA)^ 2 = (1+secA.cosecA)^2

Answer 1

(SinA +1/CosA)2 +(CosA+ 1/SinA)2      =((SinACosA +1)/CosA)2 +((SinACosA +1)/SinA)2         =(SinACosA +1)2/Cos2A +(SinACosA+1)2/Sin2A     =(SinACosA +1)2 {1/Cos2A +1/Sin2A}     =(SinACosA +1)2 {Sin2A +Cos2A /Sin2 A Cos2 A}     =(SinACosA +1)2 {1/Sin2 A Cos2 A)      =( SinA Cos A/SinA Cos A +1/SinA Cos A)2      =(1+CosecA.SecA)2      =RHS

Answer 2

$( {sinA + secA})^{2} + (cosA + cosecA) ^{2} \\ \\ = ( {sinA + \frac{1}{cosA} })^{2} + ( {cosA + \frac{1}{sinA} })^{2} \\ \\ = {sin}^{2} A + \frac{1}{ {cos}^{2} A} + 2 \frac{sinA}{cosA} + {cos}^{2} A + \frac{1}{ {sin}^{2} A} + 2 \frac{cosA}{sinA} \\ \\ = 1 + ( \frac{1}{ {sin}^{2}A } + \frac{1}{ {cos}^{2} A} ) + 2( \frac{sinA}{cosA} + \frac{cosA}{sinA} ) \\ \\ = 1 + \frac{1}{ {sin}^{2} A {cos}^{2}A } + 2 (\frac{1}{sinAcosA} ) \\ \\ = 1 + {sec}^{2} A {cosec}^{2} A + 2secAcosecA \\ \\ = ( {1 + secAcosecA})^{2}$