
If in a triangle$ABC$, the sides $AB$ and $AC$are perpendicular then the true equation is
A. $\tan A+\tan B=0$
B. $\tan B+\tan C=0$
C. $\tan A+2\tan C=0$
D. $\tan B.\tan C=1$
Answer
163.5k+ views
Hint: To solve this question, we will use the angle sum property of a triangle according to which the sum of all the angles of a triangle is $\pi $. We will derive the equation of angle $A$ in terms of angle $B$ and $C$ and then take tan on both the sides. After this we will simplify the equation using trigonometric formulas of tan and get the resultant equation.
Formula used: The trigonometric formulas of tangent are:
$\tan \left( \pi -\theta \right)=-\tan \theta $
$\tan \left( B+C \right)=\dfrac{\tan B+\tan C}{1-\tan B\tan C}$
$\tan {{90}^{0}}=\infty $
Complete step by step solution: We are given a triangle$ABC$ in which the sides $AB$ and $AC$ are perpendicular and we have to select the equation of tangent from the options.
We will take the angle sum property of the triangle and derive the equation of angle $A$ in terms of angle $B$ and $C$.
$\begin{align}
& A+B+C=\pi \\
& A=\pi -(B+C)
\end{align}$
We will now take $\tan $ on both sides of the equation.
$\tan A=\tan \left( \pi -(B+C) \right)$
As sides$AB$ and $AC$ are perpendicular the angle $A$ will be$A={{90}^{0}}$.
$\begin{align}
& \tan {{90}^{0}}=\tan \left( \pi -(B+C) \right) \\
& \infty =-\tan (B+C)
\end{align}$
We will now use the formula of $\tan \left( B+C \right)=\dfrac{\tan B+\tan C}{1-\tan B\tan C}$ and substitute it in the equation.
$\infty =-\dfrac{\tan B+\tan C}{1-\tan B\tan C}$
We will now reciprocate the above equation.
$\dfrac{1}{\infty }=-\dfrac{1-\tan B\tan C}{\tan B+\tan C}$
Now we know that any number when divided by the infinity, the resultant is always zero. So,
$\begin{align}
& 0=-\dfrac{1-\tan B\tan C}{\tan B+\tan C} \\
& 0=-\left( 1-\tan B\tan C \right) \\
& 0=\tan B\tan C-1 \\
& 1=\tan B\tan C
\end{align}$
The equation of triangle $ABC$ such that the sides $AB$ and $AC$ are perpendicular is $\tan B.\tan C=1$.
Thus, Option (D) is correct.
Note: The trigonometric ratio $\tan A$ can be defined as the ratio of the sine of the angle to the cosine of that angle. In terms of the sides of a right angled triangle, it can be defined as the ratio of the perpendicular side to the adjacent side of the triangle.
Formula used: The trigonometric formulas of tangent are:
$\tan \left( \pi -\theta \right)=-\tan \theta $
$\tan \left( B+C \right)=\dfrac{\tan B+\tan C}{1-\tan B\tan C}$
$\tan {{90}^{0}}=\infty $
Complete step by step solution: We are given a triangle$ABC$ in which the sides $AB$ and $AC$ are perpendicular and we have to select the equation of tangent from the options.
We will take the angle sum property of the triangle and derive the equation of angle $A$ in terms of angle $B$ and $C$.
$\begin{align}
& A+B+C=\pi \\
& A=\pi -(B+C)
\end{align}$
We will now take $\tan $ on both sides of the equation.
$\tan A=\tan \left( \pi -(B+C) \right)$
As sides$AB$ and $AC$ are perpendicular the angle $A$ will be$A={{90}^{0}}$.
$\begin{align}
& \tan {{90}^{0}}=\tan \left( \pi -(B+C) \right) \\
& \infty =-\tan (B+C)
\end{align}$
We will now use the formula of $\tan \left( B+C \right)=\dfrac{\tan B+\tan C}{1-\tan B\tan C}$ and substitute it in the equation.
$\infty =-\dfrac{\tan B+\tan C}{1-\tan B\tan C}$
We will now reciprocate the above equation.
$\dfrac{1}{\infty }=-\dfrac{1-\tan B\tan C}{\tan B+\tan C}$
Now we know that any number when divided by the infinity, the resultant is always zero. So,
$\begin{align}
& 0=-\dfrac{1-\tan B\tan C}{\tan B+\tan C} \\
& 0=-\left( 1-\tan B\tan C \right) \\
& 0=\tan B\tan C-1 \\
& 1=\tan B\tan C
\end{align}$
The equation of triangle $ABC$ such that the sides $AB$ and $AC$ are perpendicular is $\tan B.\tan C=1$.
Thus, Option (D) is correct.
Note: The trigonometric ratio $\tan A$ can be defined as the ratio of the sine of the angle to the cosine of that angle. In terms of the sides of a right angled triangle, it can be defined as the ratio of the perpendicular side to the adjacent side of the triangle.
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