If 0 < θ ,φ < π/2,x = Σₙ₌₀^∞ cos ²ⁿθ ,y = Σₙ₌₀^∞ sin ²ⁿφ and z = Σₙ₌₀^∞ cos ²ⁿθ .sin ²ⁿφ then :

Solution

Maths sir · 4 steps

Sum each infinite GP, write cos² θ and sin² φ in terms of x and y, then put them into z.

1. Sum each GP. For 0 < θ, φ < π/2, both cos² θ and sin² φ lie strictly between 0 and 1, so all three series converge with sum 1/(1-r). This gives x = 1/(1-cos² θ) and y = 1/(1-sin² φ).
2. Invert x and y. From these, cos² θ = 1 - 1/x and sin² φ = 1 - 1/y. The ratio of the third series is the product of these two.

   z = 1/(1-(1-1/x)(1-1/y)) = 1/(1/x+1/y-1/xy) = xy/(x+y-1)

3. Clear the fraction. Cross-multiplying gives z(x+y) - z = xy. Moving z across gives xy + z = (x+y)z.
4. So. The relation is xy + z = (x+y)z, option (c). As a check, θ = φ = π/4 gives x = y = 2 and z = 4/3, and both sides equal 16/3.

The idea to keep. When several infinite GPs share pieces, sum each one, then express the common ratios in terms of the sums. Substituting back removes the trigonometry and leaves an algebraic relation.

Answer from the NTA final answer key. Solution by PrepWiser.