Question

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

$\huge\tt\underline\orange {refer - attachment}$

$\huge\tt\underline\purple {thanks}$

$\huge\tt\underline\pink {hope-its-helpful}$

Answer 2

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \longmapsto \dfrac{sinx - cosx + 1}{sinx + cosx - 1} = \dfrac{1}{secx - tanx}$

Taking left side

Rearranging the values:

$\sf \longmapsto \dfrac{sinx + 1 - cosx}{sinx - 1 + cosx}$

$\sf \longmapsto \dfrac{sinx +(1 - cosx) }{sinx -(1 - cosx) }$

Now, Rationalising the trigonometric values:

$\sf \longmapsto \dfrac{sinx +(1 - cosx)(sinx + (1 - cosx)}{sinx - (1 -cosx) (sinx + (1 - cosx)}$

$\sf \longmapsto \dfrac{ {sin}^{2}x + (1 + {cos}^{2}x - 2cosx ) + 2sinx(1 - cosx)}{ {sin}^{2} x - (1 + {cos}^{2}x - - 2cosx) }$

$\sf \longmapsto \dfrac{ {sin}^{2} x + 1 + {cos}^{2}x - 2cosx + 2sinx - 2cosxsinx }{ {sin}^{2}x - 1 - {cos}^{2} x + 2cosx }$

$\sf \longmapsto \dfrac{1 + 1 - 2cosx + 2sinx - 2sinxcosx}{ - {cos}^{2}x - {cos}^{2}x + 2cosx }$

$\sf \longmapsto \dfrac{2 - 2cosx + 2sinx - 2sinxcosx}{ - 2 {cos}^{2} x + 2cosx}$

$\sf \longmapsto \dfrac{2(1 - cosx) + 2sinx(1 - cosx)}{2cosx(1 - cosx)}$

$\sf \longmapsto \dfrac{(2 + 2sinx)(1 - cosx)}{2cosx(1 - cosx)}$

$\sf \longmapsto \dfrac{2 + 2sinx}{2cosx}$

$\sf \longmapsto \dfrac{1 + sinx}{cosx}$

$\sf \longmapsto \: secx + tanx$

Reciprocal the term:

$\sf \longmapsto \dfrac{secx + tanx}{1}$

$\sf \longmapsto \dfrac{secx + tanx}{ {sec}^{2}x - {tan}^{2} x }$

identity used :(a+b)(a-b)=a²-b²

$\sf \longmapsto \dfrac{secx + tanx}{(secx + tanx)(secx - tanx)}$

$\sf = { \bold{ \dfrac{1}{secx - tanx} }}$

Extra shot =>

trigonometric formulas:

• Sin²θ+cos²θ=1
• sec²θ-tan²θ=1
• cosec²θ-cot²θ=1

Reciprocal Identities:

• Sinθ=1/cosecθ
• cosθ=1/secθ
• Tanθ=1/cotθ