Question

Prove that (sinA+secA) 2 +(cosA+cosecA ) 2 =(1+secAtanA ) 2

Answer 1

(sinA+secA)^2+(cosA+cosecA )^2

=(1+secAcosecA )^2

LHS

=> sin^2A + sec^2A + 2sinAsecA +

cos^2A + cosec^2A + 2cosAcosecA

=> (sin^2A + cos^2A) + 2(sinA/cosA)

+ 2(cosA/sinA) + (sec^2A + cosec^2A)

=> 1 + 2tanA + 2cotA + (sec^2A

+cosec^2A)

=> 1 + 2(tanA + cotA) + (sec^2A

+ cosec^2A)

=> 1 + 2/sinA × cosA + 1/sin^2A × cos^2A

=> (1)^2+ 2secAcosecA + (secAcosecA)^2

=> (1 + secAcosecA)^2

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Answer 2

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