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Let θ be an angle such that 

secθ = −135 and cotθ > 0
Find the exact values of tanθ and cscθ

Guest Aug 31, 2018
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By the Pythagorean identity....

 

\(\sin^2\theta+\cos^2\theta\,=\,1\\~\\ \frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}\,=\,\frac{1}{\cos^2\theta}\\~\\ \tan^2\theta+1\,=\,\sec^2\theta\)   Let's divide both sides of this equation by cos2θ, and simplify.

 

Now plug in  -135  for  sec θ   and solve the equation for  tan θ .

 

\( \tan^2\theta+1\,=\,(-135)^2\\~\\ \tan^2\theta+1\,=\,18225\\~\\ \tan^2\theta\,=\,18225-1\\~\\ \tan^2\theta\,=\,18224 \)

 

Since  cot θ  is positive,  tan θ  is positive. So take the positive square root of both sides.

 

\(\tan\theta\,=\,\sqrt{18224}\\~\\ \tan\theta\,=\,4\sqrt{1139}\)

 

And...

 

\(\frac{\sec\theta}{\tan\theta}\,=\,\sec\theta\div\tan\theta\,=\,\frac{1}{\cos\theta}\div\frac{\sin\theta}{\cos\theta}\,=\,\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}\,=\,\frac{1}{\sin\theta}\,=\,\csc\theta\)

 

So....

 

\(\csc\theta\,=\,\frac{\sec\theta}{\tan\theta}\,=\,\frac{-135}{4\sqrt{1139}}\,=\,\frac{-135\sqrt{1139}}{4556}\\~\\ \csc\theta\,=\,\frac{-135\sqrt{1139}}{4556}\)

hectictar  Aug 31, 2018

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