Question

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

Answer 1

L.H.S

(SinA + SecA )² + ( CosA + CosecA ) ²

= sin²A +sec²A +2sinAsecA +cos²A +cosec²A +2cosAcosecA

= (sin²A + cos²A) + (sec²A +cosec²A) +( 2sinA/cosA + 2cosA/sinA)

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

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

= 1 + 1/sin²Acos²A + 2/sinAcosA

= (1 + 1/sinAcosA)²

= (1 + cosecAsecA)²

Answer 2

Given,

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

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