The correct option is C ^b 1/b+1 nbsp;b+1/b 1 = 1+cosΘ+sinΘ/1+cosΘ sinΘ = 2cos^2Θ/2 + 2sinΘ/2cosΘ/2/2cos^2Θ/2 2sinΘ/2cosΘ/2 = nbsp;cosΘ/ ...

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Property 2

a=θ-tanθ, b=c...

Question

$a = sec θ - tan θ, b = cos e c θ + cot θ \Rightarrow a =$

^{\frac{b + 1}{b - 1}}

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\frac{1 + b}{1 - b}

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^{\frac{b - 1}{b + 1}}

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^{\frac{1 - b}{1 + b}}

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Solution

The correct option is C $^{\frac{b - 1}{b + 1}}$
$\frac{b + 1}{b - 1} = \frac{1 + c o s Θ + s i n Θ}{1 + c o s Θ - s i n Θ}$
$= \frac{2 c o s^{2} \frac{Θ}{2} + 2 s i n \frac{Θ}{2} c o s \frac{Θ}{2}}{2 c o s^{2} \frac{Θ}{2} - 2 s i n \frac{Θ}{2} c o s \frac{Θ}{2}}$
$= \frac{c o s \frac{Θ}{2} + s i n \frac{Θ}{2}}{c o s \frac{Θ}{2} - s i n \frac{Θ}{2}}$
$= \frac{1 + s i n Θ}{c o s Θ}$
$= \frac{c o s Θ}{1 - s i n Θ}$
$= \frac{1}{a}$

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