Prove that

Mathematics

1 answer:

Dovator [93]2 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

secx = $\frac{1}{cosx}$ , cosecx = $\frac{1}{sinx}$

cotx = $\frac{cosx}{sinx}$ , tanx = $\frac{sinx}{cosx}$

Consider the left side

secA cosecA - cotA

= $\frac{1}{cosA}$ × $\frac{1}{sinA}$ - $\frac{cosA}{sinA}$

= $\frac{1}{cosAsinA}$ - $\frac{cosA}{sinA}$

= $\frac{1-cos^2A}{cosAsinA}$

= $\frac{sin^2A}{cosAsinA}$ ( cancel sinA on numerator/ denominator )

= $\frac{sinA}{cosA}$

= tanA = right side ⇒ proven

You might be interested in

A Fair coin is tossed and a marble is chosen from the bag whose contents are shown below. what is the probability of the coin La

Vera_Pavlovna [14]

Answer:

P(Coin landing heads and blue marble is randomly selected) = 3/20.

Step-by-step explanation:

In this question, two events simultaneously take place. First, all the probabilities have to be identified. It is mentioned that the coin is a fair coin, therefore the probabilities of all the outcomes associated with the coin tossing will be equal. Therefore:

P(Coin landing heads) = 1/2.

P(Coin landing tails) = 1/2.

There are a total of 3 blue + 4 green + 3 red = 10 marbles in the bag. Therefore:

P(Selected marble is blue) = 3/10.

P(Selected marble is green) = 4/10.

P(Selected marble is red) = 3/10.

Assuming that both the events are independent, the probabilities of both the events can safely be multiplied. Therefore:

P(Coin landing heads and blue marble is randomly selected) = 1/2 * 3*10 = 3/20.

Therefore, the answer is 3/20!!!

3 0

3 years ago

Read 2 more answers