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Step-by-step explanation:
We know that
sin² θ+ cos² θ = 1
¶ cos² θ = 1 - sin² θ
→cos θ × cos θ = (1+sin θ)(1-sin θ)
→cos θ/(1-sin θ) = (1+sin θ)/cos θ
By the property of equal ratio
→cos θ/(1-sin θ) = (1+sin θ)/cos θ =
(1+sin θ-cos θ) /( cos θ- 1+ sin θ)
→cosθ/(1-sinθ) = (1+sinθ-cosθ)/(cosθ- 1+sinθ)
→cosθ/(1-sinθ) = (sinθ-cosθ+1)/(cosθ+sinθ-1)
On dividing LHS by cos θ
→(cos θ/cos θ)/(1- sin θ)/cos θ) =
(sinθ-cosθ+1)/(cosθ+sinθ-1)
→1/(1- sin θ)/cos θ) = (sinθ-cosθ+1)/(cosθ+sinθ-1)
→1/(1/Cos θ)-(Sin θ/ cos θ) =
(sinθ-cosθ+1)/(cosθ+sinθ-1)
→1/(secθ-tanθ) = (sinθ-cosθ+1)/(cosθ+sinθ-1)
→RHS = LHS
Additional comment:
If a/b = c/d then a/b = c/d
= (a+c)/(b+d)