Question

prove that (tan theata ÷ 1 - cot theata + cot theata ÷ 1 - tan theata) = 1 + sec theata - cosec theata​

Answer 1

tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)

tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1)

1/(tanθ - 1) { tan²θ - 1/tanθ }

1/(tanθ - 1) { (tan³θ - 1)/tanθ)

[as, a³ - b³ = (a - b)(a² + b² + ab)

{(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)}

tanθ + cotθ + 1

sinθ/cosθ + cosθ/sinθ + 1

(sin²θ + cos²θ)/sinθ . cosθ + 1

1/sinθ . cosθ + 1

cosecθ . secθ