Question

# 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) $\dfrac{k-1}{k+1}$

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 â€˜tanâ€™ to get the final answer.

To find the value of given expression we will write the given values first,
$\tan {{\theta }_{1}}=k\times \cot {{\theta }_{2}}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)
To proceed further we should know the formula of â€˜cot xâ€™ in terms of â€˜tan xâ€™ 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,
$\therefore \tan {{\theta }_{1}}\times \tan {{\theta }_{2}}=k$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)
Now to find the value of given expression we will first write it down and assume it as T, therefore we will get,
$T=\dfrac{\cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)}{\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€˜tanâ€™ as we have the value of $\tan {{\theta }_{1}}\times \tan {{\theta }_{2}}$.
Therefore to convert the above equation in the form of â€˜tanâ€™ 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 â€˜1â€™, therefore above equation will become,
$\therefore T=\dfrac{1-\left( \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}{1+\left( \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}} \right)}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (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 â€˜tanâ€™ otherwise the solution will become very lengthy as there will occur need of conversion of first equation.