Question

(sinA+secA)²+(cosA+cosecA)²=(1+secAcosecA)²​

Answer 1

Answer:

I assume you want me to prove this? Okay so see the picture attached for the answer.

I have used the following properties, nothing else:

$(a+b)^2 = a^2 + b^2+2ab$

$sin^2A+cos^2A=1$

$cosecA=1/sinA$

$secA=1/sinA$

$tanA=sinA/cosA$

$cotA=cosA/sinA$

Hope it helps, I have not copied from anywhere and done on my own so please mark as brainliest:)

Answer 2

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