Question

If \[\Delta = {a^2} - {(b - c)^2}\], where \[\Delta \] is the area of \[\Delta \] ABC then what is the value of \[\tan A\]?
A. \[\dfrac{{15}}{{16}}\]
B. \[\dfrac{8}{{17}}\]
C. \[\dfrac{8}{{15}}\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint: We will use the formulae of area of triangle (\[\Delta \]) and will simplfy the right-hand side expression to get the value of \[\tan \dfrac{A}{2}\]. After this we will use half angle formula to get \[\tan A\] from \[\tan \dfrac{A}{2}\].

Formula used:
Let ABC be a triangle having sides \[a\],\[b\] and \[c\] ,semi-perimeter \[s\] and area of triangle be \[\Delta \] then
\[s = \dfrac{{a + b + c}}{2}\]
\[\Delta = \sqrt {s(s - a)(s - b)(s - c)} \]
\[\tan \dfrac{A}{2} = \sqrt {\dfrac{{(s - b)(s - c)}}{{s(s - a)}}} \]
\[\tan A = \dfrac{{2\tan \dfrac{A}{2}}}{{1 - {{\tan }^2}\dfrac{A}{2}}}\]

Complete step by step solution:
Given - \[\Delta = {a^2} - {(b - c)^2}\]
Solving the RHS,
RHS = \[{a^2} - {(b - c)^2}\]
Using the formula \[{A^2} - {B^2} = (A + B)(A - B)\]
RHS = \[(a + b - c)(a - (b - c))\]
⇒ RHS= \[(a + b - c)(a - b + c)\]
Hence
 \[\Delta = (a + b - c)(a - b + c)\] …….. (1)
We know that for a triangle having sides \[a\],\[b\] and \[c\], it’s semi-perimeter \[s\] is given by formula
 \[s = \dfrac{{a + b + c}}{2}\] …….. (2)
Subtracting \[b\] from both sides of equation (2)
\[s - b = \dfrac{{a + b + c}}{2} - b\]
Simplifying the equation
\[ \Rightarrow s - b = \dfrac{{a + c - b}}{2}\]
On cross multiplying terms
\[ \Rightarrow (a + c - b) = 2(s - b)\] …….. (3)
Again, subtracting \[c\] from both sides of equation (2)
\[s - c = \dfrac{{a + b + c}}{2} - c\]
Simplifying the equation
\[ \Rightarrow s - c = \dfrac{{a + b - c}}{2}\]
On cross multiplying terms
\[ \Rightarrow (a + b - c) = 2(s - c)\] …….. (4)
By using equation (3) and (4)
\[\Delta = 2(s - b)*2(s - c)\]
\[ \Rightarrow \Delta = 4(s - b)(s - c)\] …….. (5)
Using the formula \[\Delta = \sqrt {s(s - a)(s - b)(s - c)} \] in equation (5)
\[\sqrt {s(s - a)(s - b)(s - c)} = 4(s - b)(s - c)\]
On solving the equation
\[\dfrac{{(s - b)(s - c)}}{{\sqrt {s(s - a)(s - b)(s - c)} }} = \dfrac{1}{4}\]
\[ \Rightarrow \sqrt {\dfrac{{(s - b)(s - c)}}{{s(s - a)}}} = \dfrac{1}{4}\]
But \[\tan \dfrac{A}{2} = \sqrt {\dfrac{{(s - b)(s - c)}}{{s(s - a)}}} \]
So
\[\tan \dfrac{A}{2} = \dfrac{1}{4}\] …….. (6)
Using the formula \[\tan A = \dfrac{{2\tan \dfrac{A}{2}}}{{1 - {{\tan }^2}\dfrac{A}{2}}}\] in equation (6)
\[\tan A = \dfrac{{2\left( {\dfrac{1}{4}} \right)}}{{1 - {{\left( {\dfrac{1}{4}} \right)}^2}}}\]
\[ \Rightarrow \tan A = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{16 - 1}}{{16}}}}\]
\[ \Rightarrow \tan A = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{15}}{{16}}}}\]
\[ \Rightarrow \tan A = \dfrac{8}{{15}}\]

Hence option C is correct

Note: In this question, students may get struck at the point if they try to find \[\tan A\] directly by applying the formula of \[\tan A = \dfrac{{\sin A}}{{\cos A}}\]. Here on a little simplification and manipulation of terms , we get the terms like \[s - b\],\[s - c\] and \[s\] from which we can evaluate \[\tan \dfrac{A}{2}\] and from that we can calculate the value of \[\tan A\].