3 years ago

Help mi plzz.........

1 answer:

3 0

Tan3A=tan(2A+A)

We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))

tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)

We know that , tan2x=2tanx/(1 - tan^2x)

So, by substituting tan2A in (1),we get,

=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]

=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]

=[3tanA - tan^3A]/[1-3tan^2A]

Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]

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The probability for success of an event is P(A), and the probability of success of a second, independent event is P(B). What is

Alborosie

Given:

The probability for success of an event is P(A)

The probability of success of a second, independent event is P(B).

To find:

The probability of both events occurring, in that order.

Solution:

If A and B are two independent events, then

$P(A\cap B)=P(A)\cdot P(B)$

It is given that,

The probability for success of an event is P(A)

The probability of success of a second, independent event is P(B).

Since A and B both are independent event and we need to find the probability of both events occurring, in that order, i.e., $P(A\cap B)$ , therefore $P(A\cap B)=P(A)\cdot P(B)$ .