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

Arithmetic Progression

In Δ ABC ...

Question

In $Δ$ $A B C$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively, then, $\sum \frac{b sec B + c sec C}{tan B + tan C} = ?$

2 R

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

4 R

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

6 R

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

8 R

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

Open in App

Solution

The correct option is C $6 R$
$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n B} = 2 R$

$= \frac{b c o s C + C c o s B}{s i n B - c o s C + c o s B s i n C} = \frac{b s i n B + C s i n C}{t a n B + t a n C}$

$= \frac{a}{s i n (B + C)}$
$= \frac{a}{s i n A}$

$∴ \sum \frac{a}{s i n A} = 2 R \times 3$
$= 6 R$

Suggest Corrections

Similar questions

Δ

A B C

A, B, C

a, b, c

3 a = b + c

tan \frac{B}{2} \times tan \frac{C}{2}

Δ

A B C

A, B, C

a, b, c

∠

π / 2

tan (C / 2)

Δ

A B C

A, B, C

a, b, c

a^{2}

b^{2}

c^{2}

tan A

tan B

tan C

Δ

A B C

A, B, C

a, b, c

a = 5, b = 4

tan \frac{C}{2} = \sqrt{\frac{7}{9}} .

c

Δ

A B C

A, B, C

a, b, c

A B C, \frac{tan \frac{A}{2} - tan \frac{B}{2}}{tan \frac{A}{2} + tan \frac{B}{2}}

Join BYJU'S Learning Program