Prove that

LHS = (sin θ + cosec θ)2 + (cos θ + sec θ)2
As we know,
(a + b)2 = a2 + b2 + 2ab

= sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2 cos θ sec θ
We know,
sin2θ  + cos2θ = 1, cosec2θ  = 1 + cot2θ and sec2θ = 1 + tan2θ
sin θ = 1/ cosec θ and sec θ = 1/cos θ

= 1 + 1 + cot2θ + 2 + 1 + tan2θ + 2
= 7 + tan2θ + cot2θ

= RHS

Hence proved.

OR

LHS = (1 + cot A – cosec A)(1 + tan A + sec A)
= 1(1 + tan A + sec A) + cot A (1 + tan A + sec A) - cosec A (1 + tan A + sec A)

= 1 + tan A + sec A + cot A + cot A tan A + cot A sec A – cosec A – cosec A tan A – cosec A sec A

= 2

= RHS

Hence proved.

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