sin 2θ + sin 2ϕ = 1/2

⇒ 2sin (θ + ϕ) cos (θ – ϕ) = 1/2 …(I)

cos 2θ + cos 2ϕ = 3/2

⇒ 2cos (θ + ϕ) cos (θ – ϕ) = 3/2 …(II)

Dividing equation (I) by equation (II) –

tan (θ + ϕ) = 1/3

⇒ tan² (θ +ϕ) = 1/9

⇒ 1 + tan² (θ + ϕ) = 1 + 1/9 = 10/9

Substituting this value of cos (θ + ϕ) in (II), we get –

(Ans)