The correct options are A sec^2 θ/1 cos^2 θ C sec^2 θ cosec^2 θ⇒ sec^2 θ+cosec^2 θ =1/cos^2 θ+1/sin^2 θ=sin^2 θ+cos^2 θ/cos^2 θ sin^2 θ =1/cos^2 θ sin^2 θ [ ...

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Visualisation of Trigonometric Ratios Using a Unit Circle

2θ+cosec2θ=

Question

$s e c^{2} θ + c o s e c^{2} θ =$

\frac{s e c^{2} θ}{1 - c o s^{2} θ}

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s e c^{2} θ . s i n^{2} θ

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s e c^{2} θ c o s e c^{2} θ

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\frac{c o s e c^{2} θ}{1 - c o s^{2} θ}

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Solution

The correct options are
A $\frac{s e c^{2} θ}{1 - c o s^{2} θ}$
C $s e c^{2} θ c o s e c^{2} θ$
$\Rightarrow s e c^{2} θ + c o s e c^{2} θ$

$= \frac{1}{c o s^{2} θ} + \frac{1}{s i n^{2} θ} = \frac{s i n^{2} θ + c o s^{2} θ}{c o s^{2} θ s i n^{2} θ}$

$= \frac{1}{c o s^{2} θ s i n^{2} θ}$
$[∵ s i n^{2} θ + c o s^{2} θ = 1]$

$= s e c^{2} θ c o s e c^{2} θ$

$= \frac{s e c^{2} θ}{s i n^{2} θ}$

$= \frac{s e c^{2} θ}{1 - c o s^{2} θ}$

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Similar questions

Q. Prove the following:
(iii)

s e c^{2} θ + \cos e c^{2} θ = s e c^{2} θ \times \cos e c^{2} θ

Q. If

tan θ = \frac{1}{\sqrt{7}}

then

\frac{{cosec}^{2} θ - {sec}^{2} θ}{{cosec}^{2} θ + {sec}^{2} θ} = . . . . . .

Q. If

tan θ = \frac{1}{\sqrt{3}}

then evaluate

\frac{cos e c^{2} θ - {sec}^{2} θ}{cos e c^{2} θ + {sec}^{2} θ}

Q. If

cot θ = \sqrt{3}

, find the value of

\frac{c o s e c^{2} θ + {cot}^{2} θ}{c o s e c^{2} θ - {sec}^{2} θ}

Q. If

tan θ = \frac{1}{\sqrt{7}}

, show that

\frac{c o s e c^{2} θ - {sec}^{2} θ}{c o s e c^{2} θ + {sec}^{2} θ} = \frac{3}{4}

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sec2 θ+cosec2 θ=

The correct options are A sec2 θ1−cos2 θ C sec2 θ cosec2 θ⇒sec2 θ+cosec2 θ =1cos2 θ+1sin2 θ=sin2 θ+cos2 θcos2 θ sin2 θ =1cos2 θ sin2 θ [∵sin2 θ+cos2 θ=1] =sec2 θ cosec2 θ =sec2 θsin2 θ =sec2 θ1−cos2 θ

$s e c^{2} θ + c o s e c^{2} θ =$