Question

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Answer 1

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

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$\boxed{\red{\bold{Solving \ RHS:}}}$

$\purple{\implies \displaystyle \frac{1}{sec \theta - tan \theta}}$

$\green{\boxed{\implies{\boxed{RHS = sec \theta + tan \theta}}}}$

$\boxed{\red{\bold{Solving\ LHS:}}}$

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

$\boxed{\red{\bold{Dividing \ by \ cos \theta:}}}$

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

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

$\red{\implies}\displaystyle \frac{tan \theta - 1 + sec \theta}{tan \theta + 1 - sec \theta}$

$\blue{\left( \because \displaystyle \frac{sin \theta}{cos \theta} = tan \theta \ and \ \frac{1}{cos \theta} = sec \theta \right)}$

$\red{\implies}\displaystyle \frac{tan \theta + sec \theta -1}{1 - sec \theta + tan \theta}$

$\red{\implies}\displaystyle \frac{tan \theta + sec \theta - (sec^2 \theta - tan^2 \theta)}{1-sec \theta + tan \theta}$

$\blue{(\because sec^2 \theta - tan^2 \theta = 1)}$

$\red{\implies}\displaystyle \frac{tan \theta + sec \theta -[(sec \theta + tan \theta)(sec \theta - tan \theta)]}{1 - sec \theta + tan \theta}$

$\red{\implies}\displaystyle \frac{tan \theta + sec \theta[1-(sec \theta -tan \theta)]}{1 - sec \theta + tan \theta }$

$\red{\implies}\displaystyle \frac{tan \theta + sec \theta(1-sec \theta + tan \theta)}{1- sec \theta + tan \theta}$

$\green{\boxed{\implies{\boxed{LHS = sec \theta + tan \theta}}}}$

$\Large{\overbrace{\underbrace{\boxed{ \red{H} \blue{e} \green{n} \pink{c} \purple{e} \ LHS =RHS }}}}$

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Answer 2

Step-by-step explanation:

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