Question

Prove: (1 – sinΘ) / (1 + sinΘ) = sec^2 Θ - 2secΘ tanΘ + tan^2 Θ

Answer 1

\[\dfrac{1- \sin\theta}{1 + \sin\theta} = \dfrac{(1 - \sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)} = \dfrac{1 + \sin^2\theta - 2\sin\theta}{1 - \sin^2\theta}\]\[= \dfrac{1 + \sin^2\theta - 2\sin\theta}{\cos^2\theta}\]\[= \dfrac{1}{\cos^2\theta} + \dfrac{\sin^2\theta}{\cos^2\theta } - 2\dfrac{\sin\theta}{\cos\theta}\dfrac{1}{\cos\theta}\]

Answer 2

start solving from RHS getting all terms into sin and cos terms

Answer 3

\[1/(Cos²X) – 2*SinX/(Cos²X) + (Sin²X)/(Cos²X)\]
\[(1 – 2*SinX + Sin²X)/(Cos²X)\]
\[(1-SinX)² / (1-Sin²X)\]
\[(1-SinX)*(1-SinX) /{ (1+SinX)*(1-SinX)}\]
\[(1 – SinX) / (1 + SinX) \]

Answer 4

ok i get it..thanks!