Question

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Class - 10th

Chapter - Trignometry

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Answer 1

${\huge{\rm{\underline{\underline{Question:}}}}}$

Q Prove that :

$\rm \: \frac{sin \: x - cos \: x + 1}{sin \: x + cos \: x - 1} = \frac{1}{sec \: x - {tan}^{2} \: x }$

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As , in RHS , we have tan x and sec x , so ,

in LHS we have to make it in that form ...

So ,

LHS

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

divide every term by (cos x)

$\mapsto \large\rm \: \dfrac{ \frac{sin \: x}{cos \: x} - \frac{cos \: x}{cos \: x} + \frac{1}{cos \: x} }{ \frac{sin \: x}{cos \: x} + \frac{cos \: x}{cos \: x} - \frac{1}{cos \: x} }$

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

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

Now ,

Multiplying neumerator and denominator by the conjugate of denominator ..

$\mapsto \rm \: \dfrac{(tan \: x - (1 - sec \: x))(tan \: x - (1 - sec \: x))}{(tan \: x + (1 - sec \: x))(tan \: x - (1 - sec \: x)}$

$\mapsto \rm \: \frac{ {(tan \: x - (1 - sec \: x))}^{2} }{ {tan}^{2} \: x - {(1 - sec \: x)}^{2} }$

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

$\mapsto \rm \: \frac{ {tan}^{2} \: x + 1 + {sec}^{2} \: x - 2sec \: x - 2tan \: x + 2tan \: x \: sec \: x }{ {tan}^{2} \: x - 1 - {sec}^{2} x + 2sec \: x}$

Using the Formula :

$\boxed{ \rm \: {tan}^{2} \: x + 1 = {sec}^{2} \: x} \: \\ and \\ \boxed{ \rm \: 1 = {sec}^{2} \: x - {tan}^{2} \: x}$

$\mapsto \rm \: \frac{ {sec}^{2} \: x + {sec}^{2} x - 2sec \: x - 2tan \: x + 2tan \: x \: sec \: x}{ {tan}^{2} \: x - ( {sec}^{2} \: x - {tan}^{2} \: x) - {sec}^{2} \: x + 2sec \: x }$

Now refer to the attachment

Answer 2

$\huge {\boxed {\mathbb {QUESTION}}}$

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

$\huge {\boxed {\mathbb {ANSWER}}}$

$LHS$

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

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

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

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

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

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

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

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

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

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

$RHS$

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

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$\huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$Formulas$

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

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

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

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