Question

(sinA+secA)2+(cosA+cosecA)2=(1+secAcosecA)2
Prove and explain it why we had done it like this?
Please help me

Answer 1

Given,

$(\sin A+\sec A)^{2}+(\cos A+cosecA)^{2}\\ \\=\sin^{2}A+\sec^{2}A+2\sin A\sec A+\cos^{2}A+cosec^{2}A+2\cos A.cosecA\\\\=\sin^{2}A+cos^{2}A+\sec^{2}A+2\sin A\sec A+cosec^{2}A+2\cos A.cosecA\\\\=1+\sec^{2}A+cosec^{2}A+2\sin A.\sec A+2\cos A.cosecA\\\\=1+\frac{1}{\cos^{2}A}+\frac{1}{\sin^{2}A}+\frac{2\sin A}{\cos A}+\frac{2\cos A}{\sin A}\\\\=1+\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A\cos^{2}A}+2(\frac{\sin^{2}A+\cos^{2}A}{\sin A\cos A})\\\\=1+\frac{1}{\sin^{2}A\cos^{2}A}+2(\frac{1}{\sin A\cos A})\\\\=1+cosec^{2}A.\sec^{2}A+2\times1\times\sec A.cosecA\\ \\=(1+\sec A.cosecA)^{2}\\\\\textbf{Hence,Proved}$

$Note:\\1.\sin^{2}A+\cos^{2}A=1\\\\2.\sin A=\frac{1}{cosecA}\\\\3.\cos A=\frac{1}{\sec A}\\\\4.(a+b)^{2}=a^{2}+b^{2}+ab$

Answer 2

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