Prove the following.(1) secθ (1 – sinθ) (secθ + tanθ) = 1(2) (secθ + tanθ) (1 – sinθ) = cosθ(3) sec2θ + cosec2θ = sec2θ × cosec2θ(4) cot2θ – tan2θ = cosec2θ – sec2θ(5) tan4θ + tan2θ = sec4θ – sec2θ(6) 11-sinθ+11+sinθ=2 sec2θ(7) sec6x – tan6x = 1 + 3sec2x × tan2x(8) tanθsecθ+1=secθ-1tanθ(9) tan3θ-1tanθ-1=sec2θ+tanθ(10) sinθ-cosθ+1sinθ+cosθ-1=1sinθ-tanθ

Prove the following.(1) secθ (1 – sinθ) (secθ + tanθ) = 1(2) (se

1.	Prove the following.(1) secθ (1 – sinθ) (secθ + tanθ) = 1(2) (secθ + tanθ) (1 – sinθ) = cosθ(3) sec2θ + cosec2θ = sec2θ × cosec2θ(4) cot2θ – tan2θ = cosec2θ – sec2θ(5) tan4θ + tan2θ = sec4θ – sec2θ(6) 11-sinθ+11+sinθ=2 sec2θ(7) sec6x – tan6x = 1 + 3sec2x × tan2x(8) tanθsecθ+1=secθ-1tanθ(9) tan3θ-1tanθ-1=sec2θ+tanθ(10) sinθ-cosθ+1sinθ+cosθ-1=1sinθ-tanθ
Answer» Prove the following. (1) secθ (1 – sinθ) (secθ + tanθ) = 1 (2) (secθ + tanθ) (1 – sinθ) = cosθ (3) sec²θ + cosec²θ = sec²θ × cosec²θ (4) cot²θ – tan²θ = cosec²θ – sec²θ (5) tan⁴θ + tan²θ = sec⁴θ – sec²θ (6) $\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ$ (7) sec⁶x – tan⁶x = 1 + 3sec²x × tan²x (8) $\frac{\tan θ}{s e c θ + 1} = \frac{s e c θ - 1}{\tan θ}$ (9) $\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ$ (10) $\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sin θ - \tan θ}$