If θ+φ=α And tanθ=ktanφ Then prove sinα=sin(θ-φ)(k+1)/(k-1)

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

1x2 1x3 1x4 1x5 1x6 1x7 2x1 2x2 2x3 2x4 2x5 2x6 2x7 3x1 3x2 3x3 3x4 3x5 3x6 3x7 4x1 4x2 4x3 4x4 4x5 4x6 4x7 5x1 5x2 5x3 5x4 5x5 5x6 5x8 6x1 6x2 6x3 6x4 6x5 6x6 6x7 7x1 7x2 7x3 7x4 7x5 7x6 7x7

1x2 1x3 1x4 1x5 1x6 1x7 2x1 2x2 2x3 2x4 2x5 2x6 2x7 3x1 3x2 3x3 3x4 3x5 3x6 3x7 4x1 4x2 4x3 4x4 4x5 4x6 4x7 5x1 5x2 5x3 5x4 5x5 5x6 5x8 6x1 6x2 6x3 6x4 6x5 6x6 6x7 7x1 7x2 7x3 7x4 7x5 7x6 7x7