Question

(sinA+cosecA)2 + (cosA+secA)2=tan2A+cot2A+7

Answer 1

(sinA + cosecA)² + (cosA + secA)² = tan²A + cot²A

Now,L.H.S. = (sinA + cosecA)² + (cosA + secA)²

= sin²A +cosec²A + 2 + cos²A +sec²A +2

= sin²A + cos²A + 1/sin²A + 1/cos²A + 4

= 1/sin²A + 1/cos²A + 5

= (sin²A + cos²A + 5sin²A.cos²A)/sin²A.cos²A

= ( 1 + 5 sin²A.cos²A)/sin²A.cos²A

= (sec²A.cosec²A) + 5

R.H.S. = tan²A + cot²A

= sin²A/cos²A + cos²A/ sin²A + 7

= (sin⁴A + cos⁴A)/sin²A.cos²A + 7

= {(sin²A + cos²A)² - 2sin²A.cos²A}/sin²A.cos²A + 7

= (cosec²A.sec²A)- 2 + 7

= (cosec²A.sec²A) + 5 = L.H.S. [Proved]

Silly mistake.
That's it
Hope it helped ( •̀.̫•́✧ )

Answer 2

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

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

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

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

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

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

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

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