<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csin%28x%29%20%20%2B%20%20%5Ctan%28x%29%20%7D%7B1%20%2B%20%20%5Csec%28x%29%2

All categories

2 years ago

0%7D%20%20%3D%20%20%5Csin%28x%29%20" id="TexFormula1" title=" \frac{ \sin(x) + \tan(x) }{1 + \sec(x) } = \sin(x) " alt=" \frac{ \sin(x) + \tan(x) }{1 + \sec(x) } = \sin(x) " align="absmiddle" class="latex-formula">
how do I verify this problem?

Mathematics

1 answer:

GrogVix [38]2 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

Consider the left side

Simplify the numerator/denominator using the trigonometric identities

• tanx = $\frac{sinx}{cosx}$ and secx = $\frac{1}{cosx}$

sinx + tanx = sinx + $\frac{sinx}{cosx}$ = $\frac{xsinxcosx+sinx}{cosx}$

= $\frac{sinx(1+cosx)}{cosx}$ ← numerator

and

1 + secx = 1 + $\frac{1}{cosx}$ = $\frac{cosx+1}{cosx}$ ← denominator

Putting this together gives

$\frac{sinx(1+cosx)}{cosx}$ × $\frac{cosx}{1+cosx}$

Cancel cosx and (1 + cosx) on numerator/ denominator, leaving

left side = sinx = right side ⇒ verified

You might be interested in

a coin will be tossed 10 times. Find the chance that there will be exactly 2 heads among the first five tosses and exactly 4 hea

777dan777 [17]

Answer:

The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.

Step-by-step explanation:

To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.

Both heads and tails have an individual probability p=0.5.

Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.

The probability that k heads are in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{5}{k}\cdot0.5^k\cdot0.5^{5-k}$

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

$P(x=2)=\dbinom{5}{2}\cdot0.5^{2}\cdot0.5^{3}=10\cdot0.25\cdot0.125=0.3125\\\\\\$

For the last five tosses, the probability that are exactly 4 heads is:

$P(x=4)=\dbinom{5}{4}\cdot0.5^{4}\cdot0.5^{1}=5\cdot0.0625\cdot0.5=0.1563\\\\\\$

Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:

$P(H_1=2;H_2=4)=P(H_1=2)\cdot P(H_2=4)=0.3125\cdot0.1563=0.0488$

7 0

3 years ago

Is -1 ft greater than -5ft

AfilCa [17]

Answer:

yes

Step-by-step explanation:

-1ft is greater then -5ft

3 0

2 years ago

Read 2 more answers

The table shows the prices for games in Bella's soccer league. Her parents and grandmother attended a

Kay [80]

Answer: $55

Step-by-step explanation:

You have Bella’s parents and her grandmother going to the game= 3 non student tickets at $16 per ticket. 16x3=$48 for tickets and 1 car parked = $7. $48 + $7= $55

4 0

2 years ago

Jamie needs 1/5 meter of ribbon to decorate the border of a photo frame. Enter the number of photo frames that can be decorated

charle [14.2K]

Answer:

4 photo frames 

Step-by-step explanation:

Hope it will help you 

8 0

3 years ago

Read 2 more answers

What is the 34th term of this sequence? -4, 3, 10, 17, 24

dolphi86 [110]

Answer:

238

Step-by-step explanation:

5 0

3 years ago