In a triangle,if tan A + tan B + tan C = 6 an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In any triangle $ABC$,the identity $	an A + 	an B + 	an C = 	an A 	an B 	an C$ holds true.Given $	an A + 	an B + 	an C = 6$ and $	an A \

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(C) In any triangle $ABC$,the identity $	an A + 	an B + 	an C = 	an A 	an B 	an C$ holds true.Given $	an A + 	an B + 	an C = 6$ and $	an A 	an B = 2$.Substituting these into the identity: $6 = 2 	an C \Rightarrow 	an C = 3$.Now,$	an A + 	an B = 6 - 	an C = 6 - 3 = 3$.We have the system $	an A + 	an B = 3$ and $	an A 	an B = 2$.The quadratic equation with roots $	an A$ and $	an B$ is $x^2 - 3x + 2 = 0$.Solving $(x - 1)(x - 2) = 0$,we get $x = 1$ or $x = 2$.Thus,the values are $(	an A, 	an B, 	an C) = (1, 2, 3)$ or $(2, 1, 3)$.

In a triangle,if $\tan A + \tan B + \tan C = 6$ and $\tan A \tan B = 2$,then the values of $\tan A, \tan B,$ and $\tan C$ are:

Similar Questions

If $A + B + C = 180^o,$ then the value of $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$ is

If $A+B+C=180^{\circ}$,then the value of $\tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right) \tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right) \tan \left(\frac{A}{2}\right)$ is

Let $x, y$ be positive real numbers and $m, n$ be positive integers. The maximum value of the expression $\frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}$ is

For $A, B$ and $C$,if $A+B+C=0$,then $\sin(2A) + \sin(2B) + \sin(2C)$ is equal to

The minimum value of $2 \sin^2 \theta + 8 \csc^2 \theta$ is (where $\theta \in R$):-

Vedclass Test Series

Exam Paper Generator

Online Exam Module