Question

please tell me this answer
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Answer 1

Answer:

the answer would be 1/sinθcosθ

Step-by-step explanation:

sec²θ =1/cos²θ

cosec²θ=1/sin²θ

$\sqrt{ \frac{1}{ { \cos( \alpha ) }^{2} } + \frac{1}{ \sin( { \alpha }^{2} ) } }$

Answer 2

SOLUTION:

We have,

$\sqrt{ {sec}^{2} \theta + {cosec}^{2} \theta }$

Proof:

$\sqrt{ \frac{1}{ {cos}^{2} \theta} + \frac{1}{ {sin}^{2} \theta} } \: \: \: \: \: \: \: [ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta } \: and \: {cosec}^{2} \theta = \frac{1}{ {sin}^{2} \theta } ] \\ \\ = > \sqrt{ \frac{ {sin}^{2} \theta + {cos}^{2} \theta}{ {cos}^{2} \theta {sin}^{2} \theta} } \\ \\ = > \sqrt{ \frac{1}{ {cos}^{2} \theta {sin}^{2} \theta } } \: \: \: \: \: \: \: \: \: \: \: [ {sin}^{2} \theta + {cos}^{2} \theta = 1] \\ \\ = > \frac{1}{cos \theta sin \theta} \\ \\ = > \frac{ {sin}^{2} \theta + {cos}^{2} \theta }{cos \theta sin \theta} \\ \\ = > \frac{ {sin}^{2} \theta }{cos \theta sin \theta} + \frac{ {cos}^{2} }{cos \theta sin \theta} \\ \\ = > \frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} \\ \\ = > tan \theta + cot \theta \: \: \: \: \: \: \: \: [proved]$

Thus,

The answer is tan theta + cot theta.

Hope it helps ☺️