Question

If ΔABC is an isosceles right triangle right angled at B, then $\frac{\tan A+\cot C}{\cot A+\cot C}=\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

It is given that ΔABC is an isosceles right triangle right angled at B.
$\therefore \angle B=90°\mathrm{and}\angle A=\angle C$ .....(1)
In ΔABC,
$A+B+C=180°$ (Angle sum property of a triangle)
$$\begin{gathered} \Rightarrow A+90°+C=180°\left(\mathrm{Given}\right) \\ \Rightarrow A+C=180°-90°=90° \\ \Rightarrow A=90°-C.....\left(2\right) \end{gathered}$$
$$\begin{gathered} \therefore \frac{\tan A+\cot C}{\cot A+\cot C} \\ =\frac{\tan \left(90°-C\right)+\cot C}{\cot A+\cot C}\left[\mathrm{Using}\left(2\right)\right] \\ =\frac{\cot C+\cot C}{\cot A+\cot C} \\ =\frac{2\cot C}{2\cot C}\left[\mathrm{Using}\left(1\right)\right] \end{gathered}$$
= 1

If ΔABC is an isosceles right triangle right angled at B, then $\frac{\tan A+\cot C}{\cot A+\cot C}=\underline{1}.$