Question

In a triangle $ABC$, if $A=\frac{\mathrm{\pi }}{4}\text{and}\tan B\times \tan C=k,$ then $k$ must satisfy

• A. $k^2+6k+1=0$
• B. $k^2–6k+1<0$
• C. $k^2–6k+1\ge 0$
• D. None of the above

Answer 1

The correct option is C

$k^2–6k+1\ge 0$

Determine the value of $k$

We know, that in a triangle $ABC$,

$\tan A+\tan B+\tan C=\tan A\times \tan B\times \tan C...\left(1\right)$

We know that $A=\frac{\mathrm{\pi }}{4}$ [Given]

$$\begin{gathered} \therefore \tan A=\tan \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow =1 \end{gathered}$$

Substitute the value in the equation $\left(1\right)$

$$\begin{gathered} \Rightarrow 1+\tan B+\tan C=1\times \tan B\times \tan C \\ \Rightarrow \tan B+\tan C=k-1[\because \tan B\times \tan C=k] \end{gathered}$$

We know that, for the triangle to exist$\tan B\text{and}\tan C$ must have a real solution.

We have the product and sum of roots, so we can generate a quadratic equation.

$f\left(x\right)=x^2–(k–1)x+k=0$, whose roots are $\tan B\text{and}\tan C$

The determinant of the equation is are real if $\left(b^2-4ac\right)\ge 0$, substituting the values we have,

$$\begin{gathered} \Rightarrow {(k-1)}^2-4k\ge 0 \\ \Rightarrow k^2–2k+1–4k\ge 0 \\ \Rightarrow k^2–6k+1\ge 0 \end{gathered}$$

Hence, option (C) is correct.