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Trigonometry
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Question:
Graph A:
Graph B:
Graph C:
Graph D:
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Graph E:
Graph F:
Graph G
Graph H:
Graph I:
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\[ \tan \theta~is~positive,\theta ~lies~in~1st~or~3rd~quadrant\] \[\sin \theta ~is~negative,\theta~lies~\in~3rd~or~4th~quadrant.\] \[Hence~\theta~lies~ in~3rd~quadrant.\] \[\cot \theta=\frac{ 4 }{ 3 }\] \[\csc ^2\theta-\cot ^2\theta=1\] \[\csc ^2\theta=1+\cot ^2\theta\]\[=1+(\frac{ 4 }{ 3})^2\]\[=1+\frac{ 16 }{ 9}\]\[=\frac{ 9+16 }{ 9 }\]\[=\frac{ 25 }{ 9 }\] in 3rd quadrant csc theta is negative. \[\csc \theta=-\sqrt{\frac{ 25 }{ 9}}=-\frac{ 5 }{ 3 }\] \[\sin \theta=-\frac{ 3 }{ 5 }\]
@surjithayer wrote:
\[ \tan \theta~is~positive,\theta ~lies~in~1st~or~3rd~quadrant\]
\[\sin \theta ~is~negative,\theta~lies~\in~3rd~or~4th~quadrant.\]
\[Hence~\theta~lies~ in~3rd~quadrant.\]
\[\cot \theta=\frac{ 4 }{ 3 }\]
\[\csc ^2\theta-\cot ^2\theta=1\]
\[\csc ^2\theta=1+\cot ^2\theta\]\[=1+(\frac{ 4 }{ 3})^2\]\[=1+\frac{ 16 }{ 9}\]\[=\frac{ 9+16 }{ 9 }\]\[=\frac{ 25 }{ 9 }\]
in 3rd quadrant csc theta is negative.
\[\csc \theta=-\sqrt{\frac{ 25 }{ 9}}=-\frac{ 5 }{ 3 }\]
\[\sin \theta=-\frac{ 3 }{ 5 }\]
woah
it is F
@surjithayer wrote:
it is F
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