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$\sqrt{1 + {tan}^{2}} θ \sqrt{1 + {cot}^{2} θ} \sqrt{1 - {cos}^{2} θ} \sqrt{1 - {sin}^{2} θ} =$
$sec θ$
$c o s e c θ$
$s i n θ$
1

A

c o s e c θ

B

sec θ

C

s i n θ

D

1

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Similar Questions

Q1

More than One Answer Type
एक से अधिक उत्तर प्रकार के प्रश्न

If

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

y = \frac{1 + {cos}^{2} θ c o s e c^{2} θ}{c o s e c^{2} θ} + \frac{{tan}^{2} θ}{1 + sec θ}

and

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

where 0° < θ < 90°, then the correct option(s) is/are

यदि

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

y = \frac{1 + {cos}^{2} θ c o s e c^{2} θ}{c o s e c^{2} θ} + \frac{{tan}^{2} θ}{1 + sec θ}

तथा

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

जहाँ 0° < θ < 90°, तब सही विकल्प है/हैं

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Q2

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

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Q3

IS LHS=RHS?

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ} = \frac{{sec}^{2} θ}{{tan}^{2} θ - 1}

Say true or false.

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Q4

Prove the following identity

\frac{{tan}^{2} θ}{{tan}^{2} θ - 1} + \frac{c o s e c^{2} θ}{{sec}^{2} θ - c o s e c^{2} θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ}

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Q5

Find the value of

\sqrt{1 + {tan}^{2} θ} \sqrt{1 + {cot}^{2} θ} \sqrt{1 - {cos}^{2} θ} \sqrt{1 - {sin}^{2} θ}

.
[2 Marks]

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