MCQ
If in $\text{a}\triangle\text{ABC},\tan\text{B}+\tan\text{C}=6,$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
- A$6$
- B$1$
- C$\frac16$
- DNone of these
Solution:.
In $\triangle\text{ABC},$
$\text{A+B+C}=\pi$
We know that $\tan(\text{A+B+C)}=\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}}{1-\tan\text{A}\tan\text{B}-\tan\text{B}\tan\text{C}-\tan\text{C}\tan\text{A}}$ and $\tan\pi=0.$
$\therefore\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}=0$
$\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$
If $\tan\text{A}+\tan\text{B}+\tan\text{C}=6,\tan\text{A}\tan\text{B}\tan\text{C}=6$
$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac16$
$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}=\frac16$
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The coefficient of the term independent of x in the expansion of $\Big(\text{ax}+\frac{\text{b}}{\text{x}}\Big)^{14}$ is:
$14!\ \text{a}^{7}\ \text{b}^{7}$
$\frac{14!}{7!}\ \text{a}^{7}\ \text{b}^{7}$
$\frac{14!}{(7!)^{2}}\ \text{a}^{7}\ \text{b}^{7}$
$\frac{14!}{(7!)^{3}}\ \text{a}^{7}\ \text{b}^{7}$