In Delta ABC, 1 - tan frac{A}{2}tan frac{B}{2 | vedclass

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

Question

(A) We know that in $\Delta ABC,$ $A + B + C = 180^{\circ},$ so $\frac{A}{2} + \frac{B}{2} = 90^{\circ} - \frac{C}{2}.$$1 - 	an \frac{A}{2}	an \frac

Vedclass · Accepted Answer

$\frac{2c}{a + b + c}$

Vedclass · Answer

(A) We know that in $\Delta ABC,$ $A + B + C = 180^{\circ},$ so $\frac{A}{2} + \frac{B}{2} = 90^{\circ} - \frac{C}{2}.$$1 - 	an \frac{A}{2}	an \frac{B}{2} = \frac{\cos \frac{A}{2}\cos \frac{B}{2} - \sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}$$= \frac{\cos(\frac{A}{2} + \frac{B}{2})}{\cos \frac{A}{2}\cos \frac{B}{2}} = \frac{\cos(90^{\circ} - \frac{C}{2})}{\cos \frac{A}{2}\cos \frac{B}{2}} = \frac{\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}$Using the half-angle formulas $\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}},$ $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}},$ and $\cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}},$ we get:$= \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{s(s-a)}{bc}} \cdot \sqrt{\frac{s(s-b)}{ac}}} = \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{s^2(s-a)(s-b)}{abc^2}}} = \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\frac{s}{c} \sqrt{\frac{(s-a)(s-b)}{ab}}} = \frac{c}{s}$Since $s = \frac{a+b+c}{2},$ we have $\frac{c}{s} = \frac{c}{(a+b+c)/2} = \frac{2c}{a+b+c}.$

In $\Delta ABC,$ $1 - \tan \frac{A}{2}\tan \frac{B}{2} = $

Similar Questions

In the triangle $ABC$,if $a=7, b=6$ and $A=120^{\circ}$,then the approximate value of $B$ is (in $^{\circ}$)

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

In $\triangle ABC$,$\frac{r_2(r_1+r_3)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} =$

In $\Delta ABC$, if $a^{2} \cos^{2} A - b^{2} - c^{2} = 0$, then

If the median of $\Delta ABC$ through $A$ is perpendicular to $AB$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module