Question

In triangle ABC, if sinA cosB = 1/4 and 3 tanA = tanB, then cot square A =​

Answer 1

Answer:

3

Step-by-step explanation:

In the question,

We have,

sinA.cosB = 1/4  ........(1)

and,

3tanA = tanB

So,

3(sinA/cosA) = (sinB/cosB)

3sinA.cosB = sinB.cosA

Now, on putting value of sinA.cosB in this we get,

$3 (\frac{1}{4}) =sinB.cosA\\sinB.cosA = \frac{3}{4}$

Now,

On adding this with eqn. (1) we get,

sinA.cosB + sinB.cosA = 1/4 + 3/4 = 4/4 = 1

sin(A+B) = 1 = sin 90°

So,

A+B = 90°

So,

B = 90° - A

So,

3tanA = tanB = tan(90° - A) = cotA

3(1/cotA) = cotA

So,

$cot^{2}A = 3$

Answer 2

Answer:

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