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2 years ago

Sin2A+Sin2B÷Sin2A-Sin2B=Tan(A+B)÷Tan(A-B) Prove that

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

8 0

Take L.H.S sin2A+sin2B/sin2A-sin2B

= sin2A+sin2B/sin2A-sin2B

Put

[sinC+sinD = 2sin(C+D)/2cos(C-D)/2]

[sinC-sinD = 2cos(C+D)/2.sin(C-D)/2]

= 2 sin(2A+2B)/2 cos(2A-2B)/2 / 2 cos(2A+2B) sin(2A-2B)

= sin(A+B).cos(A-B)/cos(A+B).sin(A-B)

= sin(A+B)/cos(A+B) . cos(A-B)/sin(A-B)

= tan(A+B).cot(A-B)

= tan(A+B).1/tan(A-B)

= tan(A+B)/tan(A-B)

∴ Hence we proved sin2A+sin2B/sin2A-sin2B=tan(A+B)/tan(A-B)

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A student takes a multiple-choice test that has 11 questions. Each question has five choices. The student guesses randomly at ea

marissa [1.9K]

Answer:

a) P(6) = 0.0097

b) P(More than 3) = 0.1611

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is guessed correctly, or it is not. Questions are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A student takes a multiple-choice test that has 11 questions.

This means that $n = 11$

Each question has five choices.

This means that $p = \frac{1}{5} = 0.2$

(a) Find P (6)

This is P(X = 6).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{11,6}.(0.2)^{6}.(0.8)^{5} = 0.0097$

P(6) = 0.0097

(b) Find P (More than 3).

Either P is 3 or less, or it is more than three. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 3) + P(X > 3) = 1$

We want P(X > 3). So

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{11,0}.(0.2)^{0}.(0.8)^{11} = 0.0859$

$P(X = 1) = C_{11,1}.(0.2)^{1}.(0.8)^{10} = 0.2362$

$P(X = 2) = C_{11,2}.(0.2)^{2}.(0.8)^{9} = 0.2953$

$P(X = 3) = C_{11,3}.(0.2)^{3}.(0.8)^{8} = 0.2215$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0859 + 0.2362 + 0.2953 + 0.2215 = 0.8389$

Then

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8389 = 0.1611$

P(More than 3) = 0.1611

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Sin2A+Sin2B÷Sin2A-Sin2B=Tan(A+B)÷Tan(A-B) Prove that​

Sin2A+Sin2B÷Sin2A-Sin2B=Tan(A+B)÷Tan(A-B) Prove that