If z -1-1-cos- i sin- then Re z -A- 0B- 1-2C- -2D- 1-2-2

The correct option is B 1/2 z=1/1 cosθ i sinθ z=1/1 coθ isin θ×1 coθ+i sinθ/1 coθ+i sin θ ⇒ z=1 cosθ+i sin θ/(1 cos θ)^2 (i sinθ)^2 ⇒ z=1 cosθ+i sin θ/1+cos^2 ...

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Byju's Answer

Standard XII

Mathematics

Purely Real

If z =1/1-cos...

Question

$If z = \frac{1}{1 - c o s θ - i s i n θ}, then Re(z) =$

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$\frac{1}{2}$

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$c o t \frac{θ}{2}$

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$\frac{1}{2} c o t \frac{θ}{2}$

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Solution

The correct option is B
$\frac{1}{2}$

$z = \frac{1}{1 - c o s θ - i s i n θ} z = \frac{1}{1 - c o θ - i s i n θ} \times \frac{1 - c o θ + i s i n θ}{1 - c o θ + i s i n θ} \Rightarrow z = \frac{1 - c o s θ + i s i n θ}{(1 - c o s θ)^{2} - (i s i n θ)^{2}} \Rightarrow z = \frac{1 - c o s θ + i s i n θ}{1 + c o s^{2} θ - 2 c o s θ + s i n^{2} θ} \Rightarrow z = \frac{1 - c o s θ + i s i n θ}{1 + 1 - 2 c o s θ} \Rightarrow z = \frac{1 - c o s θ + i s i n θ}{2 (1 - c o s θ)} \Rightarrow z = \frac{1 - c o s θ + i s i n θ}{2 (1 - c o s θ)} \Rightarrow Re(z) = \frac{(1 - c o s θ)}{2 (1 - c o s θ)} = \frac{1}{2}$

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