Question

Proove: sin0-cos0 +1 ÷sin0+ cos0 -1 = 1 ÷ sec0 - tan0

Answer 1

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Answer 2

∴ Hence Proved.

Step-by-step explanation:

Given,

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

LHS:

$\frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}$

$=\frac{\frac{sin\theta-cos\theta+1}{cos\theta} }{\frac{sin\theta+cos\theta-1}{cos\theta} }$ (By dividing $cos\theta$ on numerator and denominator)

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

$=\frac{(tan\theta+sec\theta)-(sec^{2}\theta-tan^{2}\theta) }{1-sec\theta+tan\theta}$

$=\frac{(tan\theta+sec\theta)-(sec\theta-tan\theta)(sec\theta+tan\theta) }{1-sec\theta+tan\theta}$

$=\frac{(tan\theta+sec\theta)(1-sec\theta+tan\theta)}{1-sec\theta+tan\theta}$

$=tan\theta+sec\theta$

$=\frac{(sec\theta+tan\theta)(sec\theta-tan\theta)}{(sec\theta-tan\theta)}$

$=\frac{(sec^{2} \theta-tan^{2} \theta)}{(sec\theta-tan\theta)}$

$=\frac{1}{sec\theta-tan\theta}=$RHS

Hence Proved.