Question

Prove LHS=RHS (SinA+cosecA)^2+(cosA+secA)^2=7+tan^2+cot^2

Answer 1

sinA+cosecA)²+(cosA+secA)²=7+tan²+cot²
LHS
=sin²A+2sinAcosecA+cosec²A+cos²A+2cosAsecA+sec²A
=(sin²A+cos²A)+2sinA*1/sinA+2cosA*1/cosA+cosec²A+sec²A
=1+2+2+1+cot²A+1+tan²A
=7+tan²A+cot²A


Hence proved


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Answer 2

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

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

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

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

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

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

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

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