Question

give solution on copy I will mark it as brainlist.. no spam answer ... class 10​

Answer 1

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Question :

$\pink\bigstar\sf{Prove \: that}$

$\quad\rm{\dfrac{sinx-cosx+1}{sinx-cosx-1} =\dfrac{1}{secx-tanx}}$

Solution :

LHS:

$\\ \quad\bigstar \large \sf \frac{sinx - cosx + 1}{sinx - cosx - 1} \\ \\ \implies \quad \large \sf \frac{sinx + (1 - cosx)}{sinx - (1 - cosx)} \\ \\ \implies \quad \large \sf \: \frac{2sin \frac{x}{2}cos \frac{x}{2} + 2 {sin}^{2} \frac{x}{2} }{2 sin \frac{x}{2}cos \frac{x}{2} - 2 {sin}^{2} \frac{x}{2} } \\ \\ \quad \implies\large \sf \ \frac{cos \frac{x}{2} + sin\frac{x}{2} }{cos \frac{x}{2} - sin \frac{x}{2} } \\ \\ \quad \implies \large \sf \: \frac{cot \frac{x}{2} - 1 }{cot \frac{x}{2} + 1 } = cot \bigg( \frac{ \pi}{4} - \frac{x}{2} \bigg) \: \\ \\ \quad \implies \large \sf \: \frac{cos \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) \times sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg)}{sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) \times sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) } \\ \\ \quad \implies \large \sf \: \frac{sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) }{1 - cos \bigg( \frac{\pi}{2} - x \bigg)} \\ \\ \quad \implies \large \sf \: \frac{cosx}{1 - sinx} \\ \\ \quad \implies \large \sf \: \frac{1}{secx - tanx}$

RHS:

Hence proved