+0  
 
0
286
1
avatar

Choose the correct answer

 Nov 12, 2018

Best Answer 

 #1
avatar+342 
+1

\(sin^2θ +cos^2θ =1\)

so \(sinθ =\sqrt{1-cos^2θ}=\sqrt{\frac{36}{36}-\frac{1}{36}}=\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{6}\)

so \(tanθ=\frac{sinθ}{cosθ}= \frac{{\frac{\sqrt{35}}{6}}}{\frac{-1}{6}}=-\sqrt{35}\)

 

Hope this helps! 

 Nov 12, 2018
 #1
avatar+342 
+1
Best Answer

\(sin^2θ +cos^2θ =1\)

so \(sinθ =\sqrt{1-cos^2θ}=\sqrt{\frac{36}{36}-\frac{1}{36}}=\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{6}\)

so \(tanθ=\frac{sinθ}{cosθ}= \frac{{\frac{\sqrt{35}}{6}}}{\frac{-1}{6}}=-\sqrt{35}\)

 

Hope this helps! 

Dimitristhym Nov 12, 2018

30 Online Users

avatar
avatar