If cos(𝜃 ) = 3/5 and tan( 𝜃 ) < 0 , find the exact value of each of the remaining trigonometric functions of 𝜃
tanθ=sinθcosθ<0\tan\theta=\frac{\sin\theta}{\cos\theta}<0tanθ=cosθsinθ<0 and cosθ=35>0\cos\theta=\frac 35>0cosθ=53>0 so we can conclude sinθ<0\sin\theta<0sinθ<0
From sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1 we can write
sinθ=−1−cos2θ=−1−9/25=−4/5\sin\theta=-\sqrt{1-\cos^2\theta}=-\sqrt{1-9/25}=-4/5sinθ=−1−cos2θ=−1−9/25=−4/5
tanθ=sinθcosθ=35:(−45)=−34\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac35:(-\frac45)=-\frac34tanθ=cosθsinθ=53:(−54)=−43
cotθ=1tanθ=−43\cot\theta=\frac{1}{\tan\theta}=-\frac43cotθ=tanθ1=−34
secθ=1cosθ=53\sec\theta=\frac{1}{\cos\theta}=\frac53secθ=cosθ1=35
cscθ=1sinθ=−54\csc\theta=\frac{1}{\sin\theta}=-\frac54cscθ=sinθ1=−45
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