Question

If $\tan {\theta _1} = k\cot {\theta _2}$, then $\dfrac{{\cos \left( {{\theta _1} - {\theta _2}} \right)}}{{\cos \left( {{\theta _1} + {\theta _2}} \right)}} =$.
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 1

Here, we will use the cosine property, $\cos \left( {{\theta _1} - {\theta _2}} \right) = \cos {\theta _1}\cos {\theta _2} + \sin {\theta _1}\sin {\theta _2}$ in the numerator and $\cos \left( {{\theta _1} + {\theta _2}} \right) = \cos {\theta _1}\cos {\theta _2} - \sin {\theta _1}\sin {\theta _2}$ in the denominator of the given equation. Then we will divide the numerator and denominator of the equation by $\cos {\theta _1}\cos {\theta _2}$ and use the tangential property $\tan {\theta _1} = \dfrac{{\sin {\theta _1}}}{{\cos {\theta _1}}}$ and $\tan {\theta _2} = \dfrac{{\sin {\theta _2}}}{{\cos {\theta _2}}}$ in the obtained equation. Then we will substitute the given value to simply it to find the required value.

Complete step by step solution: We are given that $\dfrac{{\cos \left( {{\theta _1} - {\theta _2}} \right)}}{{\cos \left( {{\theta _1} + {\theta _2}} \right)}}$.

Using the cosine property, $\cos \left( {{\theta _1} - {\theta _2}} \right) = \cos {\theta _1}\cos {\theta _2} + \sin {\theta _1}\sin {\theta _2}$ in the numerator and $\cos \left( {{\theta _1} + {\theta _2}} \right) = \cos {\theta _1}\cos {\theta _2} - \sin {\theta _1}\sin {\theta _2}$ in the denominator of the above equation, we get

$\Rightarrow \dfrac{{\cos {\theta _1}\cos {\theta _2} + \sin {\theta _1}\sin {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2} - \sin {\theta _1}\sin {\theta _2}}}$

Dividing the numerator and denominator of the above equation by $\cos {\theta _1}\cos {\theta _2}$, we get

$$\Rightarrow \dfrac{{\dfrac{{\cos {\theta _1}\cos {\theta _2} + \sin {\theta _1}\sin {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2}}}}}{{\dfrac{{\cos {\theta _1}\cos {\theta _2} - \sin {\theta _1}\sin {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2}}}}} \\\ \Rightarrow \dfrac{{\dfrac{{\cos {\theta _1}\cos {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2}}} + \dfrac{{\sin {\theta _1}\sin {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2}}}}}{{\dfrac{{\cos {\theta _1}\cos {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2}}} - \dfrac{{\sin {\theta _1}\sin {\theta _2}}}{{\cos {\theta _1}\cos {\theta _2}}}}} \\\ \Rightarrow \dfrac{{1 + \dfrac{{\sin {\theta _1}}}{{\cos {\theta _1}}} \times \dfrac{{\sin {\theta _2}}}{{\cos {\theta _2}}}}}{{1 - \dfrac{{\sin {\theta _1}}}{{\cos {\theta _1}}} \times \dfrac{{\sin {\theta _2}}}{{\cos {\theta _2}}}}} \\\$$

Using the tangential property $\tan {\theta _1} = \dfrac{{\sin {\theta _1}}}{{\cos {\theta _1}}}$ and $\tan {\theta _2} = \dfrac{{\sin {\theta _2}}}{{\cos {\theta _2}}}$ in the above equation, we get
$\Rightarrow \dfrac{{1 + \tan {\theta _1}\tan {\theta _2}}}{{1 - \tan {\theta _1}\tan {\theta _2}}}$

Substituting the value $\tan {\theta _1} = k\cot {\theta _2}$ in the above equation, we get
$\Rightarrow \dfrac{{1 + k\cot {\theta _1}\tan {\theta _2}}}{{1 - k\cot {\theta _1}\tan {\theta _2}}}$

Using the tangential property, $\cot A\tan A = 1$in the above equation, we get
$\Rightarrow \dfrac{{1 + k}}{{1 - k}}$

Hence, option A is correct.

Note:
In this problem, students need to be thorough with the basic trigonometric value of different ratios, such as sine, cosine, tangent, their properties and the ways to use them to solve it. Be careful while using the properties in the given problem as there are two angles ${\theta _1}$ and ${\theta _2}$.