The correct option is D (1 cos θ )/(1+cos θ ) (cosec θ cot θ )^2= (1 cos θ )/(1+cos θ ) L.H.S. = (cosec θ cot θ )^2 = (cosec^2θ +cot^2θ 2cosec θ cot θ ) ...

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Trigonometric Ratios

cosecθ-θ2=

Question

$(c o s e c θ - c o t θ)^{2} =$ ____________.

$\frac{(1 - c o s θ)}{(1 + c o s θ)}$

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$\frac{(1 - t a n θ)}{(1 + c o s θ)}$

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$\frac{(1 - c o s e c θ)}{(1 + c o s θ)}$

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$\frac{(1 - c o s θ)}{(1 + c o t θ)}$

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Solution

The correct option is A
$\frac{(1 - c o s θ)}{(1 + c o s θ)}$

$(c o s e c θ - c o t θ)^{2} = \frac{(1 - c o s θ)}{(1 + c o s θ)}$
$L . H . S . = (c o s e c θ - c o t θ)^{2}$
$= (c o s e c^{2} θ + c o t^{2} θ - 2 c o s e c θ c o t θ)$
$= (\frac{1}{s i n^{2} θ} + \frac{c o s^{2} θ}{s i n^{2} θ} - \frac{2 c o s θ}{s i n^{2} θ})$
$= \frac{(1 + c o s^{2} θ - 2 c o s θ)}{(1 - c o s^{2} θ)}$
$= \frac{(1 - c o s θ)^{2}}{(1 - c o s θ) (1 + c o s θ)}$
$= \frac{(1 - c o s θ)}{(1 + c o s θ)} = R . H . S .$

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Similar questions

\frac{1 + \cos θ}{1 - \cos θ} = {(cosec θ + \cot θ)}^{2}

Q. Prove that:

\frac{1 - cosθ}{1 + cosθ} = (cosecθ - cotθ)^{2} .

(i) $\frac{c o s e c θ + c o t θ}{c o s e c θ - c o t θ} = (c o s e c θ + c o t θ)^{2} = 1 + 2 c o t^{2} θ + 2 c o s e c θ c o t θ$

(ii) $\frac{s e c θ + t a n θ}{s e c θ - t a n θ} = (s e c θ + t a n θ)^{2} = 1 + 2 t a n^{2} θ + 2 s e c θ t a n θ$

Q. Prove that:

(cosec θ + cot θ)^{2} = \frac{sec θ + 1}{sec θ - 1}

Q. Solve the equation :-

\frac{1 - cos θ}{1 + cos θ} = (cosec θ - cot θ)^{2}

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(cosecθ−cotθ)2= ____________.

The correct option is A (1−cosθ)(1+cosθ) (cosecθ−cotθ)2=(1−cosθ)(1+cosθ) L.H.S.=(cosecθ−cotθ)2 =(cosec2θ+cot2θ−2cosecθcotθ) =(1sin2θ+cos2θsin2θ−2cosθsin2θ) =(1+cos2θ−2cosθ)(1−cos2θ) =(1−cosθ)2(1−cosθ)(1+cosθ) =(1−cosθ)(1+cosθ)=R.H.S.

$(c o s e c θ - c o t θ)^{2} =$ ____________.