Question

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

Answer 1

(sinA + secA)² + (cosA + cscA)² 

= (sinA + 1/cosA)² + (cosA + 1/sinA)² 

= sin²A + 2sinA/cosA + 1/cos²A + cos²A + 2cosA/sinA + 1/sin²A 

= (sin²A + cos²A) + (2sinA/cosA + 2cosA/sinA) + (1/cos²A + 1/sin²A) 

= 1 + 2(sin²A+cos²A)/(sinAcosA) + (sin²A+cos²A)/(sin²Acos²A) 

= 1 + 2/(sinAcosA) + 1/(sin²Acos²A) 

= (1 + 1/(sinAcosA))² 

= (1 + secA cscA)² 

This doesn't seem to match your solution since you have + sign after secA 
We can always go back 1 step and rewrite this differently: 

= (1 + 1/(sinAcosA))² 

= (1 + (sin²A+cos²A)/(sinAcosA))² 
= (1 + sin²A/(sinAcosA) + cos²A/(sinAcosA))² 
= (1 + sinA/cosA + cosA/sinA)² 

= (1 + tanA + cosA)² 

Answer 2

Given,

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

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