(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:
P=41.72
Step-by-step explanation:
S=ACxDB/2
81.7=8.6xDB/2
81.7=4.3xDB|:4.3
19(mm)=DB
DO=19/2=9.5
OC=8.6/2=4.3
(O is the center of the rhombus, where two diagonals meet)
a²+b²=c² (DO²+OC²=DC²)
9.5²+4.3²=c²
90.25+18,49=c²
√108,74=√c²
c≈10.43
P=4c
P=4x10.43
P=41.72
Hope it helps:)
Answer:
32
Step-by-step explanation:
So, lets go over what we know:
The equation to find x is the same on both sides.
Now, lets find this equation:
So, the total area of the two lines on the left side of the triangle are 24.
Lets first subtract the first line from the total of the two lines to find the value of the second line:
24-3
=
21.
Now that we know the value of the 2nd line is 21. lets divide it by the first line(3) to find the scale factor:
21/3
=
7
So the scale factor is 7!
Lets plug this in for x.
We know that the first line is 4.
The second line is 7x bigger than 4.
So we can calculate this as:
7x4
=
28
So x is 28!
This is your answer!
Hope it helps! :)
<span>Remember, the domain of a function is all of the x values in
the (x,y) coordinates. For this, let’s take a few bench mark coordinates. (0,0)
(1,2.5)(2,0) (3, -9). Those definitely aren’t all real numbers, so eliminate A.
X is not only less than or equal to 0, so eliminate C. </span>
<span>It’s B.</span>
Answers:
B: <em>The sine function increases on (0°, 90°) and (270°, 360°).</em>
C: <em>The cosine function decreases on (0°, 180°).</em>
E: <em>Both the cosine and sine functions have a maximum value of 1.</em>
F: <em>Both the cosine and sine functions are periodic.</em>
