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Question: If \(\cos \theta =\dfrac{-3}{5},\pi <\theta <\dfrac{3\pi }{2}\), find the values of other five trigo...

If cos⁡θ=−35,π<θ<3π2\cos \theta =\dfrac{-3}{5},\pi <\theta <\dfrac{3\pi }{2}, find the values of other five trigonometric functions and hence evaluate csc⁡θ+cot⁡θsec⁡θ−tan⁡θ\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }.

Explanation

Solution

Hint: Use the identity cos⁡2θ+sin⁡2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 to find the value of sin⁡θ\sin \theta . Hence find the values of sec⁡θ\sec \theta and csc⁡θ\csc \theta using the identities sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } respectively. Hence find the value of tan⁡θ\tan \theta and cot⁡θ\cot \theta using the identities tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and cot⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta } respectively. Hence substitute the values of csc⁡θ,cot⁡θ,sec⁡θ\csc \theta ,\cot \theta ,\sec \theta and tan⁡θ\tan \theta to get the value of csc⁡θ+cot⁡θsec⁡θ−tan⁡θ\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }.

Complete step-by-step answer:
We have cos⁡θ=−35\cos \theta =\dfrac{-3}{5}
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Substituting the value of cos⁡θ,\cos \theta , we get
sin⁡2θ+(−35)2=1⇒sin⁡2θ+925=1{{\sin }^{2}}\theta +{{\left( \dfrac{-3}{5} \right)}^{2}}=1\Rightarrow {{\sin }^{2}}\theta +\dfrac{9}{25}=1
Subtracting 925\dfrac{9}{25} from both sides, we get
sin⁡2θ=1−925=1625{{\sin }^{2}}\theta =1-\dfrac{9}{25}=\dfrac{16}{25}
Hence, we have sin⁡θ=±45\sin \theta =\pm \dfrac{4}{5}
Since π<θ<3π2\pi <\theta <\dfrac{3\pi }{2}, we have θ\theta lies in the third quadrant.
Since sine of an angle in third quadrant is negative, we have
sin⁡θ=−45\sin \theta =\dfrac{-4}{5}.
Now, we know that csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta }
Hence, we have csc⁡θ=1−45=−54\csc \theta =\dfrac{1}{\dfrac{-4}{5}}=\dfrac{-5}{4}
Also, sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }
Hence, we have sec⁡θ=1−35=−53\sec \theta =\dfrac{1}{\dfrac{-3}{5}}=\dfrac{-5}{3}.
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }
Hence, we have tan⁡θ=−45−35=43\tan \theta =\dfrac{\dfrac{-4}{5}}{\dfrac{-3}{5}}=\dfrac{4}{3}.
Using cot⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta }, we get
cot⁡θ=143=34\cot \theta =\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}
Hence, we have csc⁡θ+cot⁡θsec⁡θ−tan⁡θ=−54+34−53−43=−5+34−5−43=−24×3−9=16\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }=\dfrac{\dfrac{-5}{4}+\dfrac{3}{4}}{\dfrac{-5}{3}-\dfrac{4}{3}}=\dfrac{\dfrac{-5+3}{4}}{\dfrac{-5-4}{3}}=\dfrac{-2}{4}\times \dfrac{3}{-9}=\dfrac{1}{6}
Hence, csc⁡θ+cot⁡θsec⁡θ−tan⁡θ=16\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }=\dfrac{1}{6}

Note: Alternative solution:
Let ABC be a right-angled triangle right-angled at A. Let AC = -3 units(negative sign only describes the quadrant in which the angle lies) and BC = 5 units. Let ∠C=θ\angle C=\theta .

Now, we have
AB2+AC2=BC2⇒AB2=25−9=16⇒AB=±4A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\Rightarrow A{{B}^{2}}=25-9=16\Rightarrow AB=\pm 4
Since θ\theta lies in the third quadrant, we have AB<0AB<0
Hence, AB = -4 units.
Now, we have sin⁡θ=ABBC=−45\sin \theta =\dfrac{AB}{BC}=\dfrac{-4}{5}
Hence, we have sin⁡θ=−45\sin \theta =\dfrac{-4}{5}
Similarly, we have tan⁡θ=ABAC=−4−3=43,cot⁡θ=ACAB=34,sec⁡θ=BCAC=−53,csc⁡θ=BCAB=−54\tan \theta =\dfrac{AB}{AC}=\dfrac{-4}{-3}=\dfrac{4}{3},\cot \theta =\dfrac{AC}{AB}=\dfrac{3}{4},\sec \theta =\dfrac{BC}{AC}=\dfrac{-5}{3},\csc \theta =\dfrac{BC}{AB}=\dfrac{-5}{4}, which is the same as obtained above.
Proceeding similarly, we can find the value of csc⁡θ+cot⁡θsec⁡θ−tan⁡θ\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }.