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Componendo and Dividendo

Show that sin...

Question

Show that $\frac{sinθ - cosθ + 1}{sinθ + cosθ - 1} = \frac{1}{secθ - tanθ}$

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Solution

To prove:
$\frac{sinθ - cosθ + 1}{sinθ + cosθ - 1} = \frac{1}{secθ - tanθ}$
Proof:
$L . H . S =$ $\frac{sinθ - cosθ + 1}{sinθ + cosθ - 1}$
$=$ $\frac{(sinθ) - (cosθ - 1)}{(sinθ) + (cosθ - 1)}$ ( negative sign is taken common from the term $cosθ - 1$ )
$=$ $\frac{(sinθ) - (cosθ - 1)}{(sinθ) + (cosθ - 1)} \times \frac{(sinθ) - (cosθ - 1)}{(sinθ) - (cosθ - 1)}$ (Rationalizing the denominator)
$=$ $\frac{{[(sinθ) - (cosθ - 1)]}^{2}}{{(sinθ)}^{2} - {(cosθ - 1)}^{2}}$ $[∵ (a^{2} - b^{2}) = (a + b) (a - b)]$
$=$ $\frac{\sin^{2} θ + {(cosθ - 1)}^{2} - 2 sinθ (cosθ - 1)}{\sin^{2} θ - (\cos^{2} θ + 1 - 2 cosθ)}$ $[∵ (a^{2} - b^{2}) = (a + b) (a - b) & {(a - b)}^{2} = a^{2} + b^{2} - 2 ab]$
$=$ $\frac{\sin^{2} θ + \cos^{2} θ + {(1)}^{2} - 2 cosθ - 2 sinθcosθ + 2 sinθ}{\sin^{2} θ - \cos^{2} θ - 1 + 2 cosθ}$ $[∵ {(a - b)}^{2} = a^{2} + b^{2} - 2 ab]$
$=$ $\frac{1 + 1 - 2 cosθ - 2 sinθcosθ + 2 sinθ}{1 - \cos^{2} θ - \cos^{2} θ - 1 + 2 cosθ}$ $(∵ \sin^{2} θ + \cos^{2} θ = 1)$
$=$ $\frac{2 - 2 cosθ - 2 sinθcosθ + 2 sinθ}{- 2 \cos^{2} θ + 2 cosθ}$
$=$ $\frac{1 - cosθ - sinθcosθ + sinθ}{- \cos^{2} θ + cosθ}$ ( $2$ is cancelled because it is common in numerator and denominator.)
$=$ $\frac{(1 - cosθ) + sinθ (1 - cosθ)}{cosθ (1 - cosθ)}$ $[sinθ$ is taken common from $(- sinθcosθ + sinθ)$ term of the numerator and $cosθ$ is taken common from $(- \cos^{2} θ + cosθ)$ term of the denominator $]$
$=$ $\frac{1 + sinθ}{cosθ}$ $[$ $(1 - cosθ)$ is taken common and cancelled from the numerator and denominator $]$
$=$ $\frac{1}{cosθ} + \frac{sinθ}{cosθ}$
$=$ $secθ + tanθ$ $∵ (\frac{1}{cosθ} = secθ)$ and $(\frac{sinθ}{cosθ} = tanθ)$
$=$ $secθ + tanθ$ $\times \frac{(secθ - tanθ)}{(secθ - tanθ)}$ $[$ Multiplying the numerator and denominator with $(secθ - tanθ)$ $]$
$=$ $\frac{\sec^{2} θ - \tan^{2} θ}{(secθ - tanθ)}$
$= \frac{1}{secθ - tanθ}$ $[∵ \sec^{2} θ - \tan^{2} θ = 1]$
$= R . H . S$
Thus, $\frac{sinθ - cosθ + 1}{sinθ + cosθ - 1} = \frac{1}{secθ - tanθ}$
Hence Proved

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