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Question: If \(\csc \theta -\cot \theta =k\), then find the value of \(\tan \dfrac{\theta }{2}\)....

If csc⁡θ−cot⁡θ=k\csc \theta -\cot \theta =k, then find the value of tan⁡θ2\tan \dfrac{\theta }{2}.

Explanation

Solution

Hint: Use the fact that csc⁡2x−cot⁡2x=1{{\csc }^{2}}x-{{\cot }^{2}}x=1. Hence prove that (csc⁡x−cot⁡x)(csc⁡x+cot⁡x)=1\left( \csc x-\cot x \right)\left( \csc x+\cot x \right)=1 and hence find the value of csc⁡θ+cot⁡θ\csc \theta +\cot \theta . Hence find the value of cot⁡θ\cot \theta . Using tan⁡θ=2tan⁡θ21−tan⁡2θ2\tan \theta =\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}, form a quadratic equation in tan⁡θ2\tan \dfrac{\theta }{2} and hence find the value of tan⁡θ2\tan \dfrac{\theta }{2}. Alternatively, express csc⁡θ−cot⁡θ\csc \theta -\cot \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta . Use 1−cos⁡θ=2sin⁡2θ21-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}.

Complete step-by-step answer:
We know that csc⁡2θ−cot⁡2θ=1{{\csc }^{2}}\theta -{{\cot }^{2}}\theta =1
Using (a2−b2)=(a+b)(a−b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right), we get
(csc⁡θ−cot⁡θ)(csc⁡θ+cot⁡θ)=1\left( \csc \theta -\cot \theta \right)\left( \csc \theta +\cot \theta \right)=1
Now, we know that csc⁡θ−cot⁡θ=k (i)\csc \theta -\cot \theta =k\text{ (i)}
Hence, we have
k(csc⁡θ+cot⁡θ)=1k\left( \csc \theta +\cot \theta \right)=1
Dividing by k on both sides, we get
csc⁡θ+cot⁡θ=1k (ii)\csc \theta +\cot \theta =\dfrac{1}{k}\text{ (ii)}
Subtracting equation (ii), from equation (i), we get
2cot⁡θ=1k−k2\cot \theta =\dfrac{1}{k}-k
Hence we have 2cot⁡θ=1−k2k2\cot \theta =\dfrac{1-{{k}^{2}}}{k}
Taking reciprocals on both sides, we get
tan⁡θ2=k1−k2\dfrac{\tan \theta }{2}=\dfrac{k}{1-{{k}^{2}}}
Multiplying by 2 on both sides, we get
tan⁡θ=2k1−k2\tan \theta =\dfrac{2k}{1-{{k}^{2}}}
Now, we know that tan⁡θ=2tan⁡θ21−tan⁡2θ2\tan \theta =\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}
Hence, we have
2tan⁡θ21−tan⁡2θ2=2k1−k2\dfrac{2\tan \dfrac{\theta }{2}}{1-{{\tan }^{2}}\dfrac{\theta }{2}}=\dfrac{2k}{1-{{k}^{2}}}
Let tan⁡θ2=x\tan \dfrac{\theta }{2}=x
Hence, we have 2x1−x2=2k1−k2\dfrac{2x}{1-{{x}^{2}}}=\dfrac{2k}{1-{{k}^{2}}}
Cross multiplying, we get
2x(1−k2)=2k(1−x2) ⇒2x(1−k2)=2k−2kx2 \begin{aligned} & 2x\left( 1-{{k}^{2}} \right)=2k\left( 1-{{x}^{2}} \right) \\\ & \Rightarrow 2x\left( 1-{{k}^{2}} \right)=2k-2k{{x}^{2}} \\\ \end{aligned}
Dividing by 2 on both sides, we get
(1−k2)x=k−kx2\left( 1-{{k}^{2}} \right)x=k-k{{x}^{2}}
Adding kx2k{{x}^{2}} on both sides, we get
kx2+(1−k2)x=kk{{x}^{2}}+\left( 1-{{k}^{2}} \right)x=k
Subtracting k from both sides, we get
kx2+(1−k2)x−k=0k{{x}^{2}}+\left( 1-{{k}^{2}} \right)x-k=0
Now, we know that the roots of the quadratic expression ax2+bx+c=0a{{x}^{2}}+bx+c=0 are given by x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Here a=k,b=1−k2,c=−ka=k,b=1-{{k}^{2}},c=-k
Hence , we have x=k2−1±(1−k2)2+4k22kx=\dfrac{{{k}^{2}}-1\pm \sqrt{{{\left( 1-{{k}^{2}} \right)}^{2}}+4{{k}^{2}}}}{2k}
We know that (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}. Hence, we have
x=k2−1±1−2k2+k4+4k22k=k2−1±1+2k2+k42kx=\dfrac{{{k}^{2}}-1\pm \sqrt{1-2{{k}^{2}}+{{k}^{4}}+4{{k}^{2}}}}{2k}=\dfrac{{{k}^{2}}-1\pm \sqrt{1+2{{k}^{2}}+{{k}^{4}}}}{2k}
We know that (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}. Hence, we have
x=k2−1±(k2+1)22k=k2−1+k2+12k,k2−1−k2−12k=k,−1kx=\dfrac{{{k}^{2}}-1\pm \sqrt{{{\left( {{k}^{2}}+1 \right)}^{2}}}}{2k}=\dfrac{{{k}^{2}}-1+{{k}^{2}}+1}{2k},\dfrac{{{k}^{2}}-1-{{k}^{2}}-1}{2k}=k,\dfrac{-1}{k}
If tan⁡θ2=−1k\tan \dfrac{\theta }{2}=\dfrac{-1}{k}, we have sin⁡θ2=±11+k2,cos⁡θ2=∓k1+k2\sin \dfrac{\theta }{2}=\pm \dfrac{1}{\sqrt{1+{{k}^{2}}}},\cos \dfrac{\theta }{2}=\mp \dfrac{k}{\sqrt{1+{{k}^{2}}}}
Hence sin⁡θ=2sin⁡θ2cos⁡θ2=−2k1+k2⇒csc⁡θ=−1+k22k\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}=-\dfrac{2k}{1+{{k}^{2}}}\Rightarrow \csc \theta =-\dfrac{1+{{k}^{2}}}{2k}
Hence csc⁡θ−cot⁡θ=−1+k22k−2k1−k2=−(1−k4+4k22k(1−k2))≠1k\csc \theta -\cot \theta =-\dfrac{1+{{k}^{2}}}{2k}-\dfrac{2k}{1-{{k}^{2}}}=-\left( \dfrac{1-{{k}^{4}}+4{{k}^{2}}}{2k\left( 1-{{k}^{2}} \right)} \right)\ne \dfrac{1}{k}
Hence tan⁡θ2=−1k\tan \dfrac{\theta }{2}=\dfrac{-1}{k} is rejected.
Hence tan⁡θ2=k\tan \dfrac{\theta }{2}=k

Note: Alternative solution: Best method:
We have csc⁡θ−cos⁡θ=1−cos⁡θsin⁡θ\csc \theta -\cos \theta =\dfrac{1-\cos \theta }{\sin \theta }
We know that 1−cos⁡θ=2sin⁡2θ21-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}.
Hence csc⁡θ−cot⁡θ=2sin⁡2θ22sin⁡θ2cos⁡θ2=tan⁡θ2\csc \theta -\cot \theta =\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}=\tan \dfrac{\theta }{2}
Hence, we have
tan⁡θ2=k\tan \dfrac{\theta }{2}=k