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Question: If \(\csc \theta +\cot \theta =p\) then prove that \(\cos \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}\)...

If csc⁡θ+cot⁡θ=p\csc \theta +\cot \theta =p then prove that cos⁡θ=p2−1p2+1\cos \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}

Explanation

Solution

Hint: We have been given the value of p so we will start solving the question from RHS and then show that it is equal to LHS using the information given in question and from that we will find the value of p2{{p}^{2}} and then we substitute that value of p2{{p}^{2}} in p2−1p2+1\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1} and then by using some trigonometric formula we will show that it is equal to cos⁡θ\cos \theta .

Complete step-by-step solution -
Let’s start our solution.
We have been given csc⁡θ+cot⁡θ=p\csc \theta +\cot \theta =p
Now taking square on both the sides we get,
p2=(csc⁡θ+cot⁡θ)2{{p}^{2}}={{\left( \csc \theta +\cot \theta \right)}^{2}}
Now using the formula (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab we get,
p2=csc⁡2θ+cot⁡2θ+2csc⁡θcot⁡θ{{p}^{2}}={{\csc }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta
Now for p2−1{{p}^{2}}-1, we will use csc⁡2θ=1+cot⁡2θ{{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta and we get p2−1{{p}^{2}}-1 as,
=csc⁡2θ+cot⁡2θ+2csc⁡θcot⁡θ−1 =1+cot⁡2θ+cot⁡2θ+2csc⁡θcot⁡θ−1 =2cot⁡2θ+2csc⁡θcot⁡θ..............(1) \begin{aligned} & ={{\csc }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta -1 \\\ & =1+{{\cot }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta -1 \\\ & =2{{\cot }^{2}}\theta +2\csc \theta \cot \theta ..............(1) \\\ \end{aligned}
Now for p2+1{{p}^{2}}+1, we will use cot⁡2θ=csc⁡2θ−1{{\cot }^{2}}\theta ={{\csc }^{2}}\theta -1 and we get p2+1{{p}^{2}}+1 as,
=csc⁡2θ+cot⁡2θ+2csc⁡θcot⁡θ+1 =csc⁡2θ+csc⁡2θ−1+2csc⁡θcot⁡θ+1 =2csc⁡2θ+2csc⁡θcot⁡θ............(2) \begin{aligned} & ={{\csc }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta +1 \\\ & ={{\csc }^{2}}\theta +{{\csc }^{2}}\theta -1+2\csc \theta \cot \theta +1 \\\ & =2{{\csc }^{2}}\theta +2\csc \theta \cot \theta ............(2) \\\ \end{aligned}
Now for p2−1p2+1\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}, substituting the value of p2−1{{p}^{2}}-1 and p2+1{{p}^{2}}+1 from (1) and (2) we get,
=2cot⁡2θ+2csc⁡θcot⁡θ2csc⁡2θ+2csc⁡θcot⁡θ =2cot⁡θ(cot⁡θ+csc⁡θ)2csc⁡θ(cot⁡θ+csc⁡θ) =cot⁡θcsc⁡θ \begin{aligned} & =\dfrac{2{{\cot }^{2}}\theta +2\csc \theta \cot \theta }{2{{\csc }^{2}}\theta +2\csc \theta \cot \theta } \\\ & =\dfrac{2\cot \theta \left( \cot \theta +\csc \theta \right)}{2\csc \theta \left( \cot \theta +\csc \theta \right)} \\\ & =\dfrac{\cot \theta }{\csc \theta } \\\ \end{aligned}
Now we know that cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } using these formula we get,
=cos⁡θsin⁡θ1sin⁡θ =cos⁡θ \begin{aligned} & =\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{1}{\sin \theta }} \\\ & =\cos \theta \\\ \end{aligned}
Hence we have proved that cos⁡θ=p2−1p2+1\cos \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}.

Note: The formula that we have used cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } is very important and must be kept in mind. The most important step that we have used here was in p2−1p2+1\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1} , for p2−1{{p}^{2}}-1 we have used csc⁡2θ=1+cot⁡2θ{{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta because to cancel the number 1, and for the same reason we have used cot⁡2θ=csc⁡2θ−1{{\cot }^{2}}\theta ={{\csc }^{2}}\theta -1 for p2+1{{p}^{2}}+1 . So, this point must be clear why we are using the same formula in different ways to get the answer.