Prove- tan- -sin-tan- -sin- - -1 - -1

Applying component divide ruleLHS=D ⇒ (tanθ +sinθ )+(tanθ sinθ )(tanθ +sinθ ) (tanθ sinθ ) ⇒ 2tanθ +sinθ sinθtanθ tanθ +sinθ +sinθ ⇒ 2tanθ2sinθ =tan ...

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Standard XII

Mathematics

Parametric Differentiation

Prove: tanθ...

Question

Prove: $\frac{tan θ + sin θ}{tan θ - sin θ} = \frac{sec θ + 1}{sec θ - 1}$

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Solution

Applying component $ divide rule
$L H S = D$
$\Rightarrow \frac{(tan θ + sin θ) + (tan θ - sin θ)}{(tan θ + sin θ) - (tan θ - sin θ)}$
$\Rightarrow \frac{2 tan θ + sin θ - sin θ}{tan θ - tan θ + sin θ + sin θ}$
$\Rightarrow \frac{2 tan θ}{2 sin θ}$
$= \frac{tan θ}{sin θ} = \frac{sin θ / cos θ}{sin θ} = \frac{1}{cos θ}$
$R H S$ :
$\frac{sec θ + 1}{sec θ - 1}$
$\Rightarrow \frac{(sec θ + 1) + (sec θ - 1)}{(sec θ + 1) - (sec θ - 1)}$ [applying componendo & dividendo rule]
$= \frac{sec θ + 1 + sec θ - 1}{sec θ + 1 - sec θ + 1}$
$= \frac{2 sec θ}{2}$
$= \frac{1}{cos θ}$
$= L H S$
Hence proved

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