In triangle ABC,if x=tan left(frac{B-C}{2}rig | vedclass

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

Question

(B) In $	riangle ABC$,we use the Napier's analogy: $	an \left(\frac{B-C}{2}ight) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Given $x = 	an \left(\frac{B

Vedclass · Accepted Answer

$-xyz$

Vedclass · Answer

(B) In $	riangle ABC$,we use the Napier's analogy: $	an \left(\frac{B-C}{2}ight) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Given $x = 	an \left(\frac{B-C}{2}ight) 	an \frac{A}{2}$,substituting the analogy gives $x = \frac{b-c}{b+c} \cot \frac{A}{2} 	an \frac{A}{2} = \frac{b-c}{b+c}$.Similarly,$y = \frac{c-a}{c+a}$ and $z = \frac{a-b}{a+b}$.Now,consider $\frac{1+x}{1-x} = \frac{1 + \frac{b-c}{b+c}}{1 - \frac{b-c}{b+c}} = \frac{b+c+b-c}{b+c-b+c} = \frac{2b}{2c} = \frac{b}{c}$.Similarly,$\frac{1+y}{1-y} = \frac{c}{a}$ and $\frac{1+z}{1-z} = \frac{a}{b}$.Multiplying these,we get $\left(\frac{1+x}{1-x}ight) \left(\frac{1+y}{1-y}ight) \left(\frac{1+z}{1-z}ight) = \frac{b}{c} \cdot \frac{c}{a} \cdot \frac{a}{b} = 1$.This implies $(1+x)(1+y)(1+z) = (1-x)(1-y)(1-z)$.Expanding both sides: $1 + (x+y+z) + (xy+yz+zx) + xyz = 1 - (x+y+z) + (xy+yz+zx) - xyz$.Simplifying,we get $2(x+y+z) = -2xyz$,which means $x+y+z = -xyz$.

In $\triangle ABC$,if $x=\tan \left(\frac{B-C}{2}\right) \tan \frac{A}{2}$,$y=\tan \left(\frac{C-A}{2}\right) \tan \frac{B}{2}$,and $z=\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}$,then $(x+y+z)$ is equal to

Similar Questions

In a triangle $ABC$,$AD$ is the altitude from $A$. Given $b > c$,$\angle C = 23^o$ and $AD = \frac{abc}{b^2 - c^2}$,then $\angle B = \dots^o$.

Let $\alpha$ and $\beta$ be two distinct roots of $a \cos \theta + b \sin \theta = c$, where $a, b, c$ are three real constants and $\theta \in [0, 2\pi]$. Then, $\alpha + \beta$ is also a root of the same equation, if

In $\triangle ABC$,if $\angle C=90^{\circ}$,then $\left(\frac{r_1-r_3}{r_1}\right)\left(\frac{r_2-r_3}{r_2}\right)=$

The set of values of $a$ for which the equation $\cos\, 2x + a\, \sin\, x = 2a - 7$ possesses a solution is:

If in a $\Delta ABC$,$\cos A \cos B + \sin A \sin B \sin^2 C = 1$,then the statement which is incorrect is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module