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Answer 1

       $\underline{\bold{Step-by-step-explaination:}}$

       $\text{We must prove that:}$

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

       $\text{Taking }L.H.S$

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

       $\text{Multiplying and Dividing by }(sin\theta+cos\theta+1)$

$\implies \boxed{\frac{sin\theta - cos\theta + 1}{sin\theta + cos\theta - 1} * \frac{sin\theta + cos\theta + 1}{sin\theta + cos\theta + 1} }$

$\implies \boxed{\frac{(sin\theta + 1) - (cos\theta)}{(sin\theta + cos\theta) - (1)} * \frac{(sin\theta + 1) + (cos\theta)}{(sin\theta + cos\theta) + (1)} }$

       $\text{We know that }(a+b)(a-b) = a^{2} - b^{2}$

$\implies \boxed{\frac{(sin\theta +1)^{2} - (cos\theta)^{2}}{(sin\theta + cos\theta)^{2}- (1)^{2} } }$

       $\text{We know that }(a+b)^{2} = a^{2} + b^{2} + 2ab$

$\implies \boxed{\frac{sin^{2}\theta + 1 + 2sin\theta - cos^{2}\theta}{sin^{2}\theta + cos^{2}\theta + 2sin\theta cos\theta - 1} }$

       $\text{We know that }sin^{2}\theta + cos^{2} \theta = 1$

$\implies \boxed{\frac{sin^{2}\theta + (sin^{2}\theta + cos^{2}\theta) + 2sin\theta - cos^{2}\theta}{(1) + 2sin\theta cos\theta - 1} }$

$\implies \boxed{\frac{2sin^{2}\theta+ 2sin\theta}{2sin\theta cos\theta}}$

       $\text{Taking }2sin\theta \text{ common}$

$\implies \boxed{\frac{2sin\theta(sin\theta+1)}{2sin\theta(cos\theta)} }$

$\implies \boxed{\frac{1 + sin\theta}{cos\theta} }$

  $\therefore \text{ }L.H.S = R.H.S$

       $\text{Hence Proved}$