1. Find a. The length of side

(1+tanx)/(sinx+cosx)=secx

jek_recluse [69]

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$ is proved

<h3><u>Solution:</u></h3>

Given that,

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$ ------- (1)

First we will simplify the LHS and then compare it with RHS

$\text { L. H.S }=\frac{1+\tan x}{\sin x+\cos x}$ ------ (2)

$\text {We know that } \tan x=\frac{\sin x}{\cos x}$

Substitute this in eqn (2)

$=\frac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

On simplification we get,

$=\frac{\frac{\sin x+\cos x}{\cos x}}{\sin x+\cos x}$

$=\frac{\sin x+\cos x}{\cos x} \times \frac{1}{\sin x+\cos x}$

Cancelling the common terms (sinx + cosx)

$=\frac{1}{c o s x}$

We know secant is inverse of cosine

$=\sec x=R . H . S$

Thus L.H.S = R.H.S

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$

Hence proved

5 0

3 years ago

Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees ar

horsena [70]

Answer:

$P(A) = \frac{1}{9}$

$P(B) = \frac{1}{3}$

$P(C) = \frac{1}{729}$

Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

$P(M) = \frac{1}{3}$ .

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

$P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}$

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is $P(A) = \frac{1}{9}$ . We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

$P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}$ .

C. If the probability that two people have been hired on a specific day of the week is $3(\frac{1}{3}) ^ 2$ , then the probability that 7 people have been hired on the same day is: