(tanθ + cotθ)² = sec²θ + csc²θ
<u>Expand left side</u>: tan²θ + 2tanθcotθ + cot²θ
<u>Evaluate middle term</u>: 2tanθcotθ = = 2
⇒ tan²θ + 2+ cot²θ
= tan²θ + 1 + 1 + cot²θ
<u>Apply trig identity:</u> tan²θ + 1 = sec²θ
⇒ sec²θ + 1 + cot²θ
<u>Apply trig identity:</u> 1 + cot²θ = csc²θ
⇒ sec²θ + csc²θ
Left side equals Right side so equation is verified
Answer:
40.1% probability that he will miss at least one of them
Step-by-step explanation:
For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. 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.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
0.95 probaiblity of hitting a target
This means that
10 targets
This means that
What is the probability that he will miss at least one of them?
Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So
We want P(X < 10). So
In which
40.1% probability that he will miss at least one of them