In - ABC cot a2 - cot b2 a sin2b2 - b sin2A2-

The correct option is D c C2 [S(S a)/+S(S b)/][a(S c)(S a)/a_c+b(S b)(S c)/b_c] = S/[c]×1/c[2S^2 S(a+c+b+c)+ac+bc] =S/[2S^2 S(2S+c)+c(a+b)] ...

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

Trigonometric Ratios of Half Angles

In Δ ABC ...

Question

In $Δ$ ABC
$(c o t \frac{a}{2} + c o t \frac{b}{2})$ $(a s i n^{2} \frac{b}{2} + b s i n^{2} \frac{A}{2})$ =

cot C

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

c cot C

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

cot \frac{C}{2}

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

c cot \frac{C}{2}

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

Open in App

Solution

Suggest Corrections

Similar questions

△ A B C, (c o t \frac{A}{2} + c o t \frac{B}{2}) (a s i n^{2} \frac{B}{2} + b s i n^{2} \frac{A}{2}) =

s

A B C

(a {sin}^{2} \frac{B}{2} + b {sin}^{2} \frac{A}{2} + c)

△ A B C, (c o t \frac{A}{2} + c o t \frac{B}{2}) (a s i n^{2} \frac{B}{2} + b s i n^{2} \frac{A}{2}) =

A^{2}

B^{2}

C^{2}

Q P

c o t A

c o t B

c o t C

A . P

Δ A B C,

c o t \frac{A}{2} + c o t \frac{B}{2} + c o t \frac{C}{2}

Join BYJU'S Learning Program