Question

please answer this question ?

Class - 10th

Chapter - Trignometry

Please help me brainly users.​

Answer 1

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

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$LHS$

$\implies \frac{sin\:x-cos\:x+1}{sin\:x+cos\:x-1}$

$Divide \:each\: term \:by\: cos\: x$

$\implies \frac{\frac{sin\:x}{cos\:x}- \cancel \frac{cos\:x} {cos\:x} +\frac{1}{cos\:x}} {\frac{sin\:x}{cos\:x}+ \cancel \frac{cos\:x} {cos\:x} - \frac{1}{cos\:x}}$

$\implies \frac{tan\:x-1+sec\:x} {tan\:x+1-sec\:x}$

$\implies \frac{sec \: x + tan \: x - ( {sec}^{2}x \: - {tan}^{2}x) }{tan \: x + 1 - sec \: x}$

$\implies \frac{sec \: x + tan \: x - (sec \: x + tan \: x)(sec \: x - tan \: x)}{tan \: x + 1 - sec \: x}$

$\implies \frac{(sec \: x + tan \: x)(1 - sec \: x + tan \: x)}{tan \: x + 1 - sec \: x}$

$\implies sec \: x + tan \: x \times \frac{sec \: x - tan \: x}{sec \: x - tan \: x}$

$\implies \frac{ {sec}^{2}x - {tan}^{2} x }{sec \: x - tan \: x}$

$\implies \frac{1}{sec \: x - tan \: x}$

$RHS$

$\huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU }}}$

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$Formulas$

${sin}^{2} x+{cos}^{2}x=1$

$1+{tan}^{2}x={sex}^{2}x$

$1+{cot}^{2}x={cosec}^{2}x$

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