the minimum value of tanB*tanC in an acute triangle ABC is?

Question

a)tan(A/2),
b)cot(A/2),
c)cosec^2(A/2)
d)cot^2(A/2

please explain how to solve?

Answer 1

Set up an expression in terms of A and B, and set the partial derivative w.r.t. B equal to zero to find where the minimum value occurs for a particular value of A, then we evaluate the function at that point. Observe:

f(B)=tan B tan (π-A-B)
f'(B)=sec² B tan (π-A-B) - sec² (π-A-B) tan B

Setting this equal to zero and solving:

0=sec² B tan (π-A-B) - sec² (π-A-B) tan B
sec² (π-A-B) tan B = sec² B tan (π-A-B)

Multiplying by cos² B cos² (π-A-B)

cos² B tan B = cos² (π-A-B) tan (π-A-B)
B=π-A-B
2B=π-A
B=π/2 - A/2

Evaluating the original function at this point:

tan (π/2 - A/2) tan (π-A-(π/2 - A/2))
tan (π/2 - A/2) tan (π/2-A/2)
cot (A/2) cot (A/2)
cot² (A/2)

Therefore the answer is d) cot² (A/2)

Answer 2

That'd take a whole HEAP o' cipherin'!