Question

If in a triangle ABC, $\angle$B = 90°, then ${\tan ^2}\left( {\dfrac{A}{2}} \right)$ is:
(i)$\dfrac{{b - c}}{{b + c}}$
(ii)$\dfrac{{b + c}}{{b - c}}$
(iii)$\dfrac{{2bc}}{{b - c}}$
(iv)None of these

Answer 1

We need to find the value of ${\tan ^2}\left( {\dfrac{A}{2}} \right)$. So, for that we will be using various trigonometric identities like tan$\left( {\dfrac{{B - C}}{2}} \right)$ = $\dfrac{{b - c}}{{b + c}}\cot \left( {\dfrac{A}{2}} \right)$. First, we will calculate the value of A using the value of B = 90°. Then, we will operate tan both sides after dividing the equation by 2. We will use the above mentioned identity afterwards and upon simplification, we will get the desired value of ${\tan ^2}\left( {\dfrac{A}{2}} \right)$.

Complete step-by-step answer:
We are given $\angle$B = 90°, then A + C = B
$\Rightarrow$ A = B - C
Now, on dividing the equation both sides by 2, we get
$\dfrac{A}{2} = \dfrac{{B - C}}{2}$
After this, we will operate tan both sides in this equation. On operating tangent both sides, we get
$\Rightarrow$tan $\left( {\dfrac{A}{2}} \right)$ = tan $\left( {\dfrac{{B - C}}{2}} \right)$
And we know that tan$\left( {\dfrac{{B - C}}{2}} \right)$ = $\dfrac{{b - c}}{{b + c}}\cot \left( {\dfrac{A}{2}} \right)$
$\Rightarrow$ tan$\left( {\dfrac{A}{2}} \right)$ = $\dfrac{{b - c}}{{b + c}}\cot \left( {\dfrac{A}{2}} \right)$ = $\dfrac{{b - c}}{{b + c}}\dfrac{1}{{\tan \left( {\dfrac{A}{2}} \right)}}$
$\Rightarrow$ ${\tan ^2}\left( {\dfrac{A}{2}} \right)$ = $\dfrac{{b - c}}{{b + c}}$
Therefore, we can say that ${\tan ^2}\left( {\dfrac{A}{2}} \right)$ is equal to $\dfrac{{b - c}}{{b + c}}$i.e., option (i) is correct.

Additional Information: The tangent of a triangle is defined as: In a given right angled triangle, the tangent of an angle is calculated by the length of the side opposite to the angle divided by the length of the adjacent side of the angle. Or, in other words, we can say that the tangent of a right angled triangle can be determined by the perpendicular side of the triangle divided by the base of the triangle.

Note: In such problems where we require to use the trigonometric identities to solve the question, we need to take a good care of our basics i.e., you should know all the identities with the knowledge of when and where to use them. Such questions can be tricky and you might get confused with how to proceed further especially when there is an initial condition being provided in the question. You may get wrong while using the trigonometric identity. Also, you can solve this question by converting it into the form of: ${\tan ^2}\left( {\dfrac{A}{2}} \right) = \dfrac{{{{\sin }^2}\left( {\dfrac{A}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{A}{2}} \right)}}$ and then simplifying it.