Question

The sides $a$,$b$, $c$ of $\Delta ABC$, are in arithmetic progression. If$\cos \alpha = \dfrac{a}{{b + c}}$,$\cos \beta = \dfrac{b}{{c + a}}$, $\cos \gamma = \dfrac{c}{{a + b}}$, then ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2} = $
A.$1$
B.$\dfrac{1}{2}$
C.$\dfrac{1}{3}$
D.$\dfrac{2}{3}$

Answer 1

Hint: We are given here with the values of cosine of the angles $\alpha ,\beta ,\gamma $ and are said that the values of the sides of the triangle are in arithmetic progression and are asked to find the value of ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2}$. Thus, we will use the equations of cosine of an angle with respect to tangent of the angle and then we will proceed further.

Formulae Used:
$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

Complete step-by-step answer:
Given,
$\cos \alpha = \dfrac{a}{{b + c}}$
But, we can also write
$\cos \alpha = \dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}$
Thus, Combining the two, we can write
$\Rightarrow$ $\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} = \dfrac{a}{{b + c}}$
Then, we get
$\Rightarrow$ $\dfrac{a}{{b + c}}\left( {1 + {{\tan }^2}\dfrac{\alpha }{2}} \right) = 1 - {\tan ^2}\dfrac{\alpha }{2}$
Further,
As the values $a,b,c$ are in A.P.
Thus,
If $d$ is the common difference of the A.P.
We can say,
$b = a + d$ And $c = a + 2d$
Putting these values and proceeding, we get
$\Rightarrow$ ${\tan ^2}\dfrac{\alpha }{2} = \dfrac{{a + 3d}}{{a + b + c}} \cdot \cdot \cdot \cdot \cdot \cdot (1)$
Similarly, we get
$\Rightarrow$ ${\tan ^2}\dfrac{\gamma }{2} = \dfrac{{a - d}}{{a + b + c}} \cdot \cdot \cdot \cdot \cdot \cdot (2)$
Now,
Adding equations $(1)$ and $(2)$, we get
$\Rightarrow$ ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2} = \dfrac{{a + 3d}}{{a + b + c}} + \dfrac{{a - d}}{{a + b + c}}$
Further, we get
$\Rightarrow$ ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2} = \dfrac{{a + 3d + a - d}}{{a + b + c}}$
Then, we get
 $\Rightarrow$ ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2} = \dfrac{{2a + 2d}}{{a + b + c}}$
Substituting the values of $b$ and $c$ with respect to $a$, we get
$\Rightarrow$ ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2} = \dfrac{{2a + 2d}}{{3a + 3d}}$
Taking $(a + d)$ common in the numerator and denominator and then cancelling, we get
$\Rightarrow$ ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2} = \dfrac{2}{3}$
Hence, the correct option is (D).

Additional Information:
We can derive the formula of $\cos 2\theta $ with respect to $\tan \theta $ as per the following steps.
We know,
$\cos \left( {A + B} \right) = {\cos ^2}A - {\sin ^2}A$
Now,
For$\cos 2\theta $,
We can write,
$A = B = \theta $
Thus, we get
$\cos (\theta + \theta ) = {\cos ^2}\theta - {\sin ^2}\theta $
We can write,
$\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = {\tan ^2}\theta $
Thus, substituting this value and then taking ${\cos ^2}\theta $ common, we get
$\Rightarrow$ \[\cos 2\theta = {\cos ^2}\theta (1 - {\tan ^2}\theta )\]
Now, we can write
$\Rightarrow$ ${\cos ^2}\theta = \dfrac{1}{{{{\sec }^2}\theta }}$
Thus, we get
$\Rightarrow$ $\cos 2\theta = \dfrac{1}{{{{\sec }^2}\theta }}(1 - {\tan ^2}\theta )$
Also, we can write ${\sec ^2}\theta = 1 + {\tan ^2}\theta $
Thus, $\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

Note: We took the equation of $\cos 2\theta $ with respect to $\tan \theta $ as the question was asked about calculating a value related to the tangent of an angle. We might have also used the formula of tangent as the ratio of the sine and the cosine of the angle but for that we needed to first calculate the value of sine of the angle and then proceed which might have become clumsy. Also, we boiled down all the equations in terms of one of the three parameters $a,b,c$ so as to simplify it further. We took in terms of $a$. But if the student wants, he or she can take the values in terms of $b$ or $c$ also.