GONZALEZ

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \right) \\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } }{ \cos ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

3 years ago

I will mark you Brainliest just help me

Alika [10]

Answer:

The first one is proportional and the second one is not. Because for the first one you can see that if you divide the number of pens by the cost you'll get the same answer for every single one. But for the second one if you divide the number of minutes by words typed you get a different answer for each one. You can also see if it's proportional or not if you draw it on the graph and it's a straight line. If it is the relationship is proportional if it isn't it's not proportional. Hope that makes sense :)

7 0

3 years ago

Read 2 more answers

Stephanie is hosting an art show and she would like to raise at least $800. She plans on

Masteriza [31]

Answer: 10, 15

Step-by-step explanation:

Given

Stephanie wants to raise at least $800

she plans on selling VIP tickets for $20 each and general admission for $40 each

suppose there are x and y tickets for VIP and general admission

$\therefore x+y\leq 25\\\\\Rightarrow 20x+40y\geq 800$

solving above two inequality

Stephanie must sell 10 tickets for VIP and 15 tickets for general admission

7 0

3 years ago

Answer the following questions:

timama [110]

Answer:

Question 1: 76 dollars. ($76)

2. 1/5

3. 1/19

4. 1/5

Step-by-step explanation:

QUESTION 1:

228 (the amount Mrs. Nicci had) divided by 3 (children) is equal to 76

228/3 = 76

QUESTIONS 2, 3, AND 4.

2. Find the greatest common divisor (GCD). the GCD of 14 and 70 is 14. Then, divide the numerator and denominator by 14.

14/14 = 1

70/14 = 5

WHICH EQUALS 1/5!!

3. Find the greatest common divisor (GCD). the GCD of 5 and 95 is 5. Divide both the numerator and denominator by 5.

5/5= 1

95/5= 19

WHICH EQUALS 1/19!!

4. Find the greatest common divisor (GCD). the GCD of 12 and 60 is 12. So (like we have been doing for the whole part of this question), divide both the numerator and denominator by 12, which is the GCD.

12/12 = 1

60/12 = 5

WHICH EQUALS 1/5!!

Hope this helped.. I think this is like one of the longest Brainly answers I've recently had to type lol

4 0

3 years ago

The probability of spinning an even number, flipping tails, then spinning an odd number is

nlexa [21]

Answer:

Step-by-step explanation:

6 0

2 years ago