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Question: In a triangle ABC, \(\Delta = {a^2} - {\left( {b - c} \right)^2}\) then find the value of \(\tan A\)...

In a triangle ABC, Δ=a2−(b−c)2\Delta = {a^2} - {\left( {b - c} \right)^2} then find the value of tan⁡A\tan A

Explanation

Solution

In the given question triangle equation is given, we have to apply the suitable formula according to the question to find the value of tan A. First apply the formula to find the perimeter of the triangle which is s=a+b+c2s = \dfrac{{a + b + c}}{2} where s is the semi perimeter of the triangle and a, b, c are sides of the triangle. We also apply the formula a=2s(b+c)a = 2s\left( {b + c} \right).Put both formulas in the given equation and solve it. We also use the formula of tan⁡A2\tan \dfrac{A}{2} to find the value oftan⁡A\tan A. Thus we get the correct answer.
Formula:
s=a+b+c2 and a=2s−(b+c)  s = \dfrac{{a + b + c}}{2} \\\ and \\\ a = 2s - \left( {b + c} \right) \\\

Complete step by step answer:

Given that:
Δ=a2−(b−c)2\Delta = {a^2} - {\left( {b - c} \right)^2}
Put the formulas in the given equation:
We get:
Δ=[2s−(b+c)2]−(b−c)2\Delta = \left[ {2s - {{\left( {b + c} \right)}^2}} \right] - {\left( {b - c} \right)^2}
Using the formula (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
Δ=[4s2+(b+c)2−4s(b+c)]−(b−c)2 ⇒4s2−4s(b+c)+4bc ⇒4s2−4sb+4sc+4bc ⇒4s(s−b)−4c(s−b) ⇒(4s−4c)(s−b) ⇒Δ=4(s−c)(s−b) ⇒14=(s−c)(s−b)Δ..............(1)  \Delta = \left[ {4{s^2} + {{\left( {b + c} \right)}^2} - 4s\left( {b + c} \right)} \right] - {\left( {b - c} \right)^2} \\\ \Rightarrow 4{s^2} - 4s\left( {b + c} \right) + 4bc \\\ \Rightarrow 4{s^2} - 4sb + 4sc + 4bc \\\ \Rightarrow 4s\left( {s - b} \right) - 4c\left( {s - b} \right) \\\ \Rightarrow \left( {4s - 4c} \right)\left( {s - b} \right) \\\ \Rightarrow \Delta = 4\left( {s - c} \right)\left( {s - b} \right) \\\ \Rightarrow \dfrac{1}{4} = \dfrac{{\left( {s - c} \right)\left( {s - b} \right)}}{\Delta }..............\left( 1 \right) \\\
We know that the formula for
tan⁡A2=(s−b)(s−c)s(s−a)\tan \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}}
Where s is the semi parameter of the triangle
Now from the formula:
We get:
(s−b)(s−c)=tan⁡A2s(s−a)   \sqrt {\left( {s - b} \right)} \left( {s - c} \right) = \tan \dfrac{A}{2}\sqrt {s\left( {s - a} \right)} \\\ \\\
Multiplying both sides by (s−b)(s−c)   \sqrt {\left( {s - b} \right)} \left( {s - c} \right) \\\ \\\
(s−b)(s−c)=tan⁡A2s(s−a)(s−b)(s−c) ⇒(s−b)(s−c)=tan⁡A2Δ ⇒(s−b)(s−c)Δ=tan⁡A2.........(2)  \left( {s - b} \right)\left( {s - c} \right) = \tan \dfrac{A}{2}\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \\\ \Rightarrow \left( {s - b} \right)\left( {s - c} \right) = \tan \dfrac{A}{2}\Delta \\\ \Rightarrow \dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{\Delta } = \tan \dfrac{A}{2}.........\left( 2 \right) \\\
By using both equations 1&2
We get:
14=tan⁡A2 ⇒tan⁡A=2tan⁡A21−tan⁡2A  \dfrac{1}{4} = \tan \dfrac{A}{2} \\\ \Rightarrow \tan A = \dfrac{{2\tan \dfrac{A}{2}}}{{1 - {{\tan }^2}A}} \\\
This is the formula for tan A
Now put the values

tan⁡A=2(14)1−(14)2 ⇒815  \tan A = \dfrac{{2\left( {\dfrac{1}{4}} \right)}}{{1 - {{\left( {\dfrac{1}{4}} \right)}^2}}} \\\ \Rightarrow \dfrac{8}{{15}} \\\

Hence we get the value of tan A i.e. 815\dfrac{8}{{15}}

Note: First of all remember all the trigonometric formulas especially used in these types of questions. We have to learn and remember all the useful concepts and formulas. Then apply the formulas according to the given question, put the values very carefully, and calculate the answer. In this manner, we will get the correct answer.