Question

(sinA+cosecA)^2 + (cosA+secA)^2 = tan^2A + cot^2A + 7

Answer 1

hope this is helpful.. :)

Answer 2

$\bf\huge LHS = (sinA + cosecA)^2 + (cosA + secA)^2$

$\bf\huge (sin^2 A + cosec^2 A + 2sinA cosecA ) + (cos^2 A + sec^2 A + 2 cosA SecA)$

$\bf\huge (sin^2 A + cosec^2 A + 2 sinA . \frac{1}{sinA}) + (cos^2A + sec^2 A + 2 cosA . \frac{1}{cosA})$

$\bf\huge (sin^2 A + cosec^2 A + 2) + (cos^2 A + sec^2 A + 2)$

$\bf\huge (sin^2 A + cos^2 A + 2) + (cos^2 A + sec^2 A + 2)$

$\bf\huge sin^2 A + cos^2 A + cosec^2 A + sec^2 A + 4$

$\bf\huge 1 + (1 + cot^2) + (1 + tan^2 A) + 4$

$\bf\huge 7 + tan^2 A + cot^2 A$