Question

PLEASE HELP AWARDING METAL!!

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

1 + sec2x sin2x = sec2x

Answer 1

left side:
we can replace sec x with 1/cos x
so we can write:

\[1 + {\left( {\sec x} \right)^2}{\left( {\sin x} \right)^2} = 1 + \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}}\]

Answer 2

How?

Answer 3

here are more steps:
\[1 + {\left( {\sec x} \right)^2} \times {\left( {\sin x} \right)^2} = 1 + \frac{1}{{{{\left( {\cos x} \right)}^2}}} \times {\left( {\sin x} \right)^2} = 1 + \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}}\]

Answer 4

what is the first step?

Answer 5

now we can write this:
\[1 + \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = \frac{{{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 6

please simplify

Answer 7

I don't have much time!

Answer 8

you have to use the fundamental identity of trigonometry, namely:
\[{\left( {\cos x} \right)^2} + {\left( {\sin x} \right)^2} = 1\]

Answer 9

so we can write:
\[1 + \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = \frac{{{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = \frac{1}{{{{\left( {\cos x} \right)}^2}}}\]

Answer 10

ok this is the first step...

Answer 11

right side:
we can rewrite the right side, as below:
\[{\left( {\sec x} \right)^2} = \frac{1}{{{{\left( {\cos x} \right)}^2}}}\]

Answer 12

so we have shown that left side is equal to right side, and our work is completed

Answer 13

same problem, different question.
- tan2x + sec2x = 1

Answer 14

here we have to replace tan x with its definition, namely tan x = sin x/ cos x, and we have to replace sec x with its definition, namely:
sec x= 1/ cos x, so we get:
\[ - {\left( {\tan x} \right)^2} + {\left( {\sec x} \right)^2} = - \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} + \frac{1}{{{{\left( {\cos x} \right)}^2}}} = ...?\]
please continue

Answer 15

please give me one minute.

Answer 16

ok!

Answer 17

Thank you so much for your Patience :)

Answer 18

thanks! :)

Answer 19

could you go over it and explain after.

Answer 20

yes I can, nevertheless you also have to contribute in getting your answers

Answer 21

next step is:
\[ - \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} + \frac{1}{{{{\left( {\cos x} \right)}^2}}} = \frac{{1 - {{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 22

Ok, sorry time is just working against me here.

Answer 23

please keep in mind that I have to respect the Code of Conduct

Answer 24

now we have:
\[ - \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} + \frac{1}{{{{\left( {\cos x} \right)}^2}}} = \frac{{1 - {{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = \frac{{{{\left( {\cos x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 25

Do I simplify?

Answer 26

yes!

Answer 27

hint:

Answer 28

−(sinx)2/(cosx)2+1/(cosx)2=1−(sinx)2/(cosx)2=

Answer 29

yes! and the next step is:
\[\frac{{1 - {{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = \frac{{{{\left( {\cos x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 30

so, what do you get?

Answer 31

now we have:
−(sinx)2/(cosx)2+1/(cosx)2=1−(sinx)2/(cosx)2= 1-(sinx)^2/ (cosx)^2

Answer 32

Right?

Answer 33

yes! it is right! nevertheless you have to simplify this expression:
\[\frac{{{{\left( {\cos x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 34

is the result 1?

Answer 35

yes

Answer 36

ok! so we have shown that the left side is equal to 1, which is the right side, then our work is completed

Answer 37

please wait a moment I have to answer to my phone

Answer 38

Ok.

Answer 39

1*1/ (sinx)^2/ (cosx)^2

Answer 40

here I am

Answer 41

Right?

Answer 42

can you write your expression with the editor, since I don't understand it

Answer 43

or please make a drawing of your expression

Answer 44

is your expression like this?
\[\Large \frac{{1 \times 1}}{{{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2}}}\]

Answer 45

Yes, is it right?

Answer 46

are you referring to the second exercise?

Answer 47

nevermind...

Answer 48

the first two identities have been checked
have you another exercise to solve?

Answer 49

yes! one last one. Give me one minute.

Answer 50

ok!

Answer 51

@clairvoyant1

Answer 52

hey sorry, let us continue.

Answer 53

ok!

Answer 54

what is your question?

Answer 55

@clairvoyant1

Answer 56

Please, if you need help with your mathematics, then tag me @clairvoyant1

Answer 57

\[\frac{ sinx }{ 1-cosx }+\frac{ sinx }{ 1+cosx }=2 \csc x\]

Answer 58

@Michele_Laino

Answer 59

@welshfella