In triangle ABC-sin A-sin B-sin Csin A-sin B-sin C is equal to

We know that nbsp; A+B+C = π Now, sin A+sin B sin C =2sinA+B2cosA B2 2sinC2cosC2 (∵sin(A+B2) = sin(π2 C 2 ) = cos C 2 ) =2cosC2[cosA B2 sinC2] =2cosC2[co ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Conditional Identities

In triangle A...

Question

In triangle $A B C, \frac{sin A + sin B + sin C}{sin A + sin B - sin C}$ is equal to

tan \frac{A}{2} cot \frac{B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cot \frac{A}{2} tan \frac{B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cot \frac{A}{2} cot \frac{B}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

tan \frac{A}{2} tan \frac{B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $cot \frac{A}{2} cot \frac{B}{2}$
We know that
$A + B + C = π$
Now,
$sin A + sin B - sin C = 2 sin \frac{A + B}{2} cos \frac{A - B}{2} - 2 sin \frac{C}{2} cos \frac{C}{2} (∵ sin (\frac{A + B}{2}) = sin (\frac{π}{2} - \frac{C}{2}) = cos \frac{C}{2}) = 2 cos \frac{C}{2} [cos \frac{A - B}{2} - sin \frac{C}{2}] = 2 cos \frac{C}{2} [cos \frac{A - B}{2} - cos \frac{A + B}{2}] (∵ sin (\frac{C}{2}) = sin (\frac{π}{2} - \frac{A + B}{2}) = cos \frac{A + B}{2}) = 4 sin \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2}$

Similarly
$sin A + sin B + sin C = 2 cos \frac{C}{2} [cos \frac{A - B}{2} + cos \frac{A + B}{2}] = 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}$

$∴ \frac{sin A + sin B + sin C}{sin A + sin B - sin C} = cot \frac{A}{2} cot \frac{B}{2}$

Suggest Corrections

Similar questions

Q. In a triangle

A B C

, if

(sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B

is equal to

In $△ A B C, R r (sin A + sin B + sin C)$ is equal
to

Q. In a triangle ABC, if

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0

then

sin A sin B + sin B sin C + sin C sin A

is equal to

Q. In

△ A B C, \frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C}

is equal to :-

Q. In a

△ A B C,

{sin}^{2} A - {sin}^{2} B + {sin}^{2} C

is equal to

Join BYJU'S Learning Program

In triangle ABC,sinA+sinB+sinCsinA+sinB−sinC is equal to

In triangle $A B C, \frac{sin A + sin B + sin C}{sin A + sin B - sin C}$ is equal to