Question

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

Answer 1

This is not 5 points question, because this is easy but yeah not that easy,

first of all lets do few things in advance
(1)
$\frac{1}{sin^{2}x} + \frac{1}{cos^{2}x} \\\\= \frac{sin^{2}x+cos^{2}x}{sin^{2}xcos^{2}x} \\ \\ = \frac{1}{sin^{2}xcos^{2}x}$

(2)
$\frac{sinx}{cosx} + \frac{cosx}{sinx} \\ \\ = \frac{sin^{2}x + cos^{2}x}{sinxcosx} \\ \\ \frac{1}{sinxcosx}$

Proof;
$(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}$$[sin^{2}A + cos^{2}A] + [\frac{1}{sin^{2}A} + \frac{1}{cos^{2}A}] + 2[\frac{sinA}{cosA} + \frac{cosA}{sinA}]$$1 + \frac{1}{sin^{2}Acos^{2}A} + \frac{2}{sinAcosA} \\ \\ (1 + \frac{1}{sinAcosA})^{2} \\ \\ (1 + secAcosecA)^{2}$

Answer 2

Given,

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

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