Question

Pls prove it using $ identity sec*2theta = 1+ tan*2theta.

Answer 1

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Answer 2

Question :-- Prove that (sin A-cos A+1)/(sin A+cosA-1)=1/(sec A-tan A)

Formula used :---

• 1 = sec²A - tan²A = (secA + TanA)
• SinA/cosA = TanA
• 1/CosA = SecA

Solution :---

Dividing the L.H.S. Numerator and denominator by cos A we get,

→ (tan A-1+secA)/(tan A+1-sec A)

→ (tan A-1+secA)/(1-sec A+tan A)

Now , Putting 1 = (sec A+tan A)(secA-tanA) in Denominator , we get,

→(sec A+tan A-1)/[(sec A+tan A)(sec A-tan A)-(sec A-tan A)]

Taking (secA-tanA) common From Denominator now ,

→ (sec A+tan A-1)/(sec A-tan A)(sec A+tan A-1)

→ 1/(sec A-tan A) = R.H.S

✪✪ Hence Proved ✪✪