Question

The sides $a,b,c$ of $\vartriangle ABC$, are in A.P. 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

Since, the sides of the triangle are in A.P, let $d$ be the common difference of A.P and rewrite the sides as $a = b - d$, $b$ and $c = b + d$. Then, substitute the values in the given equation $\cos \alpha = \dfrac{a}{{b + c}}$, $\cos \beta = \dfrac{b}{{c + a}}$, $\cos \gamma = \dfrac{c}{{a + b}}$ and simplify using the trigonometric formulas. Further, find the value of ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2}$.

Complete step-by-step answer:
We are given that the sides of a triangle are in A.P
Let $d$ be the common difference of A.P
Then, the sides of A.P. can be written as $a = b - d$, $b$ and $c = b + d$
Also, we are given than $\cos \alpha = \dfrac{a}{{b + c}}$
Substitute the value of $a,b,c$ in the equation.
$\cos \alpha = \dfrac{{b - d}}{{b + b + d}}$
$\Rightarrow \cos \alpha = \dfrac{{b - d}}{{2b + d}}$ (1)
Similarly, we will write the value of $\cos \beta = \dfrac{b}{{c + a}}$
Hence, we get,
$\cos \beta = \dfrac{b}{{b + d + b - d}} \\\ \Rightarrow \cos \beta = \dfrac{b}{{2b}} \\\ \Rightarrow \cos \beta = \dfrac{1}{2} \\\ \Rightarrow \beta = {60^ \circ } \\\$
Now, we will write the value of $\cos \gamma = \dfrac{c}{{a + b}}$
$\cos \gamma = \dfrac{c}{{a + b}} \\\ \Rightarrow \cos \gamma = \dfrac{{b + d}}{{b - d + b}} \\\$
$\Rightarrow \cos \gamma = \dfrac{{b + d}}{{2b - d}}$ (2)
Now, we know that $\cos \left( {2x} \right) = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
Therefore from equation (1), we have,
$\cos \alpha = \dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} \\\ \Rightarrow \dfrac{{b - d}}{{2b + d}} = \dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} \\\$
We will solve the above equation to find the value of ${\tan ^2}\dfrac{\alpha }{2}$
Use componendo and dividendo rules to simplify the equation.
$\dfrac{{b - d - 2b - d}}{{b - d + 2b + d}} = \dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2} - 1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 - {{\tan }^2}\dfrac{\alpha }{2} + 1 + {{\tan }^2}\dfrac{\alpha }{2}}} \\\ \Rightarrow \dfrac{{ - b - 2d}}{{3b}} = \dfrac{{ - 2{{\tan }^2}\dfrac{\alpha }{2}}}{2} \\\ \Rightarrow {\tan ^2}\dfrac{\alpha }{2} = \dfrac{{2d + b}}{{3b}} \\\$
Similarly, we will find the value of ${\tan ^2}\dfrac{\gamma }{2}$ from equation (2)
$\Rightarrow \cos \gamma = \dfrac{{1 - {{\tan }^2}\dfrac{\gamma }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}} \\\ \Rightarrow \dfrac{{b + d}}{{2b - d}} = \dfrac{{1 - {{\tan }^2}\dfrac{\gamma }{2}}}{{1 + {{\tan }^2}\dfrac{\gamma }{2}}} \\\$
Use componendo and dividendo rules to simplify the equation.
$\dfrac{{b + d - 2b + d}}{{b + d + 2b - d}} = \dfrac{{1 - {{\tan }^2}\dfrac{\gamma }{2} - 1 - {{\tan }^2}\dfrac{\gamma }{2}}}{{1 - {{\tan }^2}\dfrac{\gamma }{2} + 1 + {{\tan }^2}\dfrac{\gamma }{2}}} \\\ \Rightarrow \dfrac{{ - b + 2d}}{{3b}} = \dfrac{{ - 2{{\tan }^2}\dfrac{\gamma }{2}}}{2} \\\ \Rightarrow {\tan ^2}\dfrac{\gamma }{2} = \dfrac{{b - 2d}}{{3b}} \\\$
We will now substitute the values of ${\tan ^2}\dfrac{\alpha }{2}$ and ${\tan ^2}\dfrac{\gamma }{2}$ in the expression ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2}$
$\dfrac{{2d + b}}{{3b}} + \dfrac{{b - 2d}}{{3b}} = \dfrac{{2d + b + b - 2d}}{{3b}} \\\ \Rightarrow \dfrac{{2b}}{{3b}} \\\ \Rightarrow \dfrac{2}{3} \\\$
Hence, the value of ${\tan ^2}\dfrac{\alpha }{2} + {\tan ^2}\dfrac{\gamma }{2}$ is $\dfrac{2}{3}$.
Thus, option D is correct.

Note: Here, we have used the formula of $\cos \left( {2x} \right) = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$, where the angle gets half. Similarly, we have $\cos \left( {2x} \right) = {\cos ^2}x - {\sin ^2}x$, $\cos \left( {2x} \right) = 1 - 2{\sin ^2}x$ and $\cos \left( {2x} \right) = 2{\cos ^2}x - 1$. Use the formula according to the condition in the question.