In ∆ABC, the value of1−tan⁡B2tan⁡C2 is

1−tan⁡B2⋅tan⁡C2=1−(s−a)(s−c)s(s−b)⋅(s−a)(s−b)s(s−c) =1−s−as=s−s+as=as=aa+b+c2=2aa+b+c

In ∆ABC, the value of1−tan⁡B2tan⁡C2 is

1−tan⁡B2⋅tan⁡C2=1−(s−a)(s−c)s(s−b)⋅(s−a)(s−b)s(s−c) =1−s−as=s−s+as=as=aa+b+c2=2aa+b+c

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In $∆ A B C$ , the value of $1 - \tan \frac{B}{2} \tan \frac{C}{2}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

$\frac{4 a}{a + b + c}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$1 - \tan \frac{B}{2} \cdot \tan \frac{C}{2} = 1 - \sqrt{\frac{(s - a) (s - c)}{s (s - b)}} \cdot \sqrt{\frac{(s - a) (s - b)}{s (s - c)}} \begin{array}{l} = 1 - \frac{s - a}{s} = \frac{s - s + a}{s} \\ = \frac{a}{s} = \frac{a}{\frac{a + b + c}{2}} = \frac{2 a}{a + b + c} \end{array}$