Question

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⇒ Prove that sinφ-cosφ+1 / sinφ+cosφ-1 = 1 / secφ-tanφ

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¥ ~ reGarDs

Answer 1

Hey there !!


$\large{ \boxed{ \boxed{ \mathbb{TRIGONOMETRY.}}}}$



Prove that :-

$\frac{sin \theta - cos \theta + 1}{sin \theta + cos \theta - 1} = \frac{1}{sec \theta - tan \theta}$

For Solution :-


$\boxed{ \boxed{ \bf{See \: the \: attachment.}}}$


▶ Identity used :-

→ sin ∅/ cos ∅ = tan ∅ .

→ 1/ cos ∅ = sec ∅ .

→ sec²∅ - tan²∅ = 1 .

→ a² - b² = ( a + b ) ( a - b ) .



$\huge \bf \underline{ \mathbb{LHS = RHS .}}$


✔✔ Hence, it is proved ✅✅.

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THANKS



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Answer 2

Welcome Friend :)

Here's your answer !

sinθ-cosθ+1 / sinθ+1 (LHS)

On dividing numerator and denominator by cosθ, we get

= sinθ/cosθ - cosθ/cosθ + 1/cosθ

 sinθ/cosθ + cosθ/cosθ - 1/cosθ

=  tanθ+secθ-1

   tanθ-secθ+1

=(sec²θ-tan²θ)

  tanθ+1-secθ

= (tanθ+secθ)-[(secθ+tanθ) (secθ-tanθ)

  tanθ+1-secθ

= (tanθ+secθ) (1-secθ+tanθ)

 = tanθ+secθ

 = secθ+tanθ ....... (1)

RHS = 1 / secθ- tanθ

        = 1 / secθ-tanθ x secθ+tanθ/secθ+tanθ

by rationalising RF of secθ-tanθ is secθ+tanθ

= secθ+tanθ / sec²θ-tan²θ

= secθ+tanθ ....... (2)

From 1 and 2, we get

LHS = RHS

hence proved *

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