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Question

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

A

${k}^{2}+6k+1=0$

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B

${k}^{2}–6k+1<0$

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C

${k}^{2}–6k+1\ge 0$

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D

None of the above

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Solution

## The correct option is C ${k}^{2}–6k+1\ge 0$Determine the value of $k$We know, that in a triangle $ABC$,$\mathrm{tan}A+\mathrm{tan}B+\mathrm{tan}C=\mathrm{tan}A×\mathrm{tan}B×\mathrm{tan}C...\left(1\right)$We know that $A=\frac{\mathrm{\pi }}{4}$ [Given]$\therefore \mathrm{tan}A=\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒=1$Substitute the value in the equation $\left(1\right)$$⇒1+\mathrm{tan}B+\mathrm{tan}C=1×\mathrm{tan}B×\mathrm{tan}C\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}B+\mathrm{tan}C=k-1\left[\because \mathrm{tan}B×\mathrm{tan}C=k\right]$We know that, for the triangle to exist$\mathrm{tan}B\text{and}\mathrm{tan}C$ must have a real solution.We have the product and sum of roots, so we can generate a quadratic equation.$f\left(x\right)={x}^{2}–\left(k–1\right)x+k=0$, whose roots are $\mathrm{tan}B\text{and}\mathrm{tan}C$The determinant of the equation is are real if $\left({b}^{2}-4ac\right)\ge 0$, substituting the values we have,$⇒{\left(k-1\right)}^{2}-4k\ge 0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}–2k+1–4k\ge 0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}–6k+1\ge 0$Hence, option (C) is correct.

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