Question

prove that (sinA+cosecA)^2+(cosA+secA)^2 equal to 7+tan^2A+cot^2A

Answer 1

=(sinA+cosecA)²+(cosA+secA)²

=sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA

=sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA

=1+cosec²A+sec²A+2+2

=5+(1+cot²A)+(1+tan²A)

=7+tan²A+cot²A

Identities used:
1+tan²A=sec²A

1+cot²A=cosec²A

sin²A+cos²A=1

cosecA=1/sinA

secA=1/cosA

Answer 2

Hi !

Here's the answer to your query

LHS = (sin A + cosec A)2 + (cos A+ sec A)2

= (sin2 A + 2 sin A . cosec A + cosec2 A) + (cos2 A + 2 cos A . sec A + sec2 A)

= sin2 A + 2 + cosec2 A + cos2 A + 2 + sec2 A ---(sin A . cosec A =1=cos A . sec A)

= sin2 A + 2 + 1 + cot2 A + cos2 A + 2 + 1 + tan2 A ---(cosec2 A=1+ cot2 A & sec2 A=1+tan2 A)

= ( sin2 A + cos2 A ) + 6 + ( cot2 A + tan2 A )

= 1 + 6 + cot2 A + tan2 A-----(sin2 A + cos2 A =1)

= 7 + tan2 A + cot2

= RHS

Hence proved

Cheers!!!