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Find the value of following expression, If $\tan {{\theta }_{1}}=k\times \cot {{\theta }_{2}}$:
\[\dfrac{\cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)}{\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}\]
A) \[\dfrac{1+k}{1-k}\]
B) \[\dfrac{1-k}{1+k}\]
C) \[\dfrac{k+1}{k-1}\]
D) \[\dfrac{k-1}{k+1}\]

Answer
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Hint: For solving this question use the formulae \[\cos \left( A+B \right)=\cos A\times \cos B-\sin A\times \sin B\] and \[\cos \left( A-B \right)=\cos A\times \cos B+\sin A\times \sin B\] to expand the given expression and convert it in the form of ‘tan’ to get the final answer.

To find the value of given expression we will write the given values first,
$\tan {{\theta }_{1}}=k\times \cot {{\theta }_{2}}$…………………………….. (1)
To proceed further we should know the formula of ‘cot x’ in terms of ‘tan x’ which is given below,
Formulae:
\[\cot x=\dfrac{1}{\tan x}\]
By using above formula we can write the equation (1) as,
$\therefore \tan {{\theta }_{1}}=k\times \dfrac{1}{\tan {{\theta }_{2}}}$
If we shift $\tan {{\theta }_{2}}$ on the right hand side of the equation we will get,
$\therefore \tan {{\theta }_{1}}\times \tan {{\theta }_{2}}=k$ …………………………………….. (2)
Now to find the value of given expression we will first write it down and assume it as T, therefore we will get,
\[T=\dfrac{\cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)}{\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}\] ………………………………… (A)
Now to solve further we should know the formulae given below,
Formulae:
\[\cos \left( A+B \right)=\cos A\times \cos B-\sin A\times \sin B\]
\[\cos \left( A-B \right)=\cos A\times \cos B+\sin A\times \sin B\]
By using above formulae we can write T as shown below,
\[\therefore T=\dfrac{\left( \cos {{\theta }_{1}}\times \cos {{\theta }_{2}}-\sin {{\theta }_{1}}\times \sin {{\theta }_{2}} \right)}{\left( \cos {{\theta }_{1}}\times \cos {{\theta }_{2}}+\sin {{\theta }_{1}}\times \sin {{\theta }_{2}} \right)}\]
If we observe above equation and equation (2) then we can say that we have to convert above equation in the form of ‘tan’ as we have the value of \[\tan {{\theta }_{1}}\times \tan {{\theta }_{2}}\].
Therefore to convert the above equation in the form of ‘tan’ we will divide both numerator and denominator by \[\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}\] and therefore we will get,
\[\therefore T=\dfrac{\left( \dfrac{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}-\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}{\left( \dfrac{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}+\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}\]
By separating the denominators of numerator and denominator we will get,
\[\therefore T=\dfrac{\left( \dfrac{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)-\left( \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}{\left( \dfrac{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)+\left( \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}\]
As we all know that if any term is divided by itself we will get the answer equal to ‘1’, therefore above equation will become,
\[\therefore T=\dfrac{1-\left( \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}{1+\left( \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}\] …………………………………….. (3)
Now, to proceed further in the solution we should know the formula given below,
Formula:
\[\dfrac{\sin x}{\cos x}=\tan x\]
By using above formula we can write the equation (3) as shown below,
\[\therefore T=\dfrac{1-\tan {{\theta }_{1}}\times \tan {{\theta }_{2}}}{1+\tan {{\theta }_{1}}\times \tan {{\theta }_{2}}}\]
If we put the value of equation (2) in above equation we will get,
\[\therefore T=\dfrac{1-k}{1+k}\]
If we compare above equation with equation (A) we will get,
\[\therefore T=\dfrac{\cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)}{\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}=\dfrac{1-k}{1+k}\]
\[\therefore \dfrac{\cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)}{\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}=\dfrac{1-k}{1+k}\]
Therefore the value of \[\dfrac{\cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)}{\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}\] is given b, \[\dfrac{1-k}{1+k}\] .
Therefore Option (b) is correct.

Note- Do remember the adjustment of dividing the expression by \[\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}\] after expansion using formula to convert it in to the form of ‘tan’ otherwise the solution will become very lengthy as there will occur need of conversion of first equation.