Question

Prove that: (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2A.​

Answer 1

Step-by-step explanation:

=(sinA+cosecA)²+(cosA+secA)²

=sin²A+cosec²A+2sinAcosecA+cos²A+se

=sin²A+cos²A+cosec²A+sec²A+2sinAx1/:

=1+cosec²A+sec²A+2+2

=5+(1+cot²A)+(1+tan²A)

=7+tan²A+cot²A

Answer 2

(sin A + cos e * c * A) ^ 2 + (cos A + sec(A)) ^ 2 = 7 + tan^2 A + cot^2 A

L.H.S.

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

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

2 cos A. sec A

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

= 1+ 2+2+ (cot² A + 1) + (tan² A + 1)

= 5 + 1 + 1 + cot² A + tan² A

= 7+ tan² A + cot² A = R.H.S.

Hence, L.H.S. = R.H.S.