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STatiana [176]
3 years ago
7

Given cos theta = - 2/5 and sin theta < 0, find the six trigonometric values. I need help with this

Mathematics
1 answer:
Lena [83]3 years ago
3 0

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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Answer:

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4 0
3 years ago
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In exponent multiplication, if base are same just add the powers.

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