3 years ago

The probability can be written as: Fraction Decimal Percent All of the above

Mathematics

1 answer:

Alla [95]3 years ago

5 0

All of the above. Sorry if this is wrong but in 99% sure it isn’t!!

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Answer:

$\sin\,\theta =-\frac{\sqrt{21} }{5}$

$\tan\,\theta =\frac{\sqrt{21} }{2}$

$\sec\,\theta = \frac{-5}{2}$

$cosec\,\theta =\frac{-5}{\sqrt{21} }$

$\cot\,\theta =\frac{2}{\sqrt{21} }$

Step-by-step explanation:

$\cos\theta =\frac{-2}{5}$

As both $sin\,\theta$ ,

$\theta$ lies in the third quadrant.

In the third quadrant,

$\sin\theta$

$\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}$

$\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}$

$\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}$

$\ cosec \,\theta = \frac{1}{sin\,\theta }\\\\=\frac{1}{\frac{-\sqrt{21} }{5} }\\\\=\frac{-5}{\sqrt{21} }$

$\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }$

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lara31 [8.8K]

Zero
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