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sinθ- cosθ+1 / sinθ+cosθ-1 = 1/secθ-tanθ​

Sagot :

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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)

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