Question

Guys help me on Trigonometric Identities......

Answer 1

I am here only, just post your question.... :P

Answer 2

Alright...

Answer 3

Reciprocal Identity

Answer 4

Alright??

Sorry. this isn't any identity I have studied in Trigonometry???

Answer 5

Yes, what about Reciprocal Identity??

Answer 6

\[\sec^2\theta-1/\sec^2\]

Answer 7

Okay, what to do with it??

Answer 8

How do solve for that?

Answer 9

I neeed to prove its equal to cos^2theta tan^2theta

Answer 10

Do you mean this:

\[\frac{1 - \sec^2(x)}{\sec^2(x)}\]

Answer 11

Do you know this Identity :

\[1 + \tan^2(x) = \sec^2(x)\]

Answer 12

yes. pythagorean identity....

Answer 13

Then use this, you will get that..

Answer 14

See, here : \(1 + tan^2(x) = sec^2(x)\)

Subtract 1 from both the sides :

\(tan^2(x) = sec^2(x) - 1\) Getting this step??

Answer 15

\[\tan^2\theta/\sec^2\]

Answer 16

One step away.. Use now reciprocal Identity for sec(x)... :)

Answer 17

am i right?

Answer 18

Pretty much right, you are going pretty accurate than me... :)

Answer 19

Im getting all confused actually....

Answer 20

Where and why and how??

Answer 21

because of the denominator.....

Answer 22

Reciprocal Identity for sec(x) says that : \(\large sec(x) = \frac{1}{cos(x)}\)

Answer 23

See, divide it into simpler parts, so that you can look it in a better way:

can I write this expression as :

\[\tan^2(x) \cdot \frac{1}{\sec(x)} \cdot \frac{1}{\sec(x)}\]

Answer 24

Now you know by reciprocal identity that cos(x) = 1/sec(x)

Answer 25

Still not getting @EngrChong

Answer 26

so it proves they're equal?

Answer 27

Okay, I need to go by another method:

You got upto here : \(\large \frac{tan^2(x)}{sec^2(x)}\) Right??

Answer 28

yeah

Answer 29

See, you are just solving the Left Hand Side, if from Left Hand Side you can make Right Hand Side, that means the two are equal...

Answer 30

Do you know cos(x) = 1/sec(x) ??

Answer 31

How is that?

Answer 32

oh yeah

Answer 33

See, here (sec^2-1)/sec^2(x), is this your left hand side or right hand side??

Answer 34

lefthand side

Answer 35

So, if you simplify your left hand side and finally make it look like same as What you are given on right hand side that means RHS and LHS are equal and hence you have proved this identity..

Answer 36

Can you replace there 1/sec^2(x) with cos^2(x) so that you can get your right hand side??

Answer 37

Wait im a little confused wih your question.....

Answer 38

In tan^2(x)/sec^2(x) what you can do more in this to achieve your right hand side??

Answer 39

the sec^2(x) should be changed into 1/cos^2(x)

Answer 40

So you mean like this:

\[\large \frac{\tan^2(x)}{\frac{1}{\cos^2(x)}}\]

Okay, I agree with you..
And finally what will you get now??

Answer 41

yes so 1/cos^2(x) becomes cos^2(x)

Answer 42

What??

Answer 43

You mean that cos^2(x) will come into numerator and will join tan^2(x) his best buddy hanging out there?? "P

Answer 44

well the right hand side is \[\cos^2\theta \tan^2\theta\]

Answer 45

Why are you telling me what is right hand side???

Make your LHS look like RHS..

Answer 46

I am telling you what will be your answer to this:

\[\large \frac{\tan^2(x)}{\frac{1}{\cos^2(x)}}\]

Answer 47

See,

\[\large \frac{a}{\frac{b}{c}} = \frac{ac}{b}\]

Answer 48

Use this concept there..

Answer 49

So it's \[\tan^2\theta \cos^2\theta \]

Answer 50

Yes, this you have simplified your LHS like this. Don't you think that It is matching to RHS??

If yes, then you have proved your Identity..

Answer 51

Similarly if you make your RHS look like your LHS, then also we say Identity has been proved..

I am doing it fro you:

Our RHS is:

\[\tan^2(x) \cos^2(x)\]

Using Reciprocal Identity as : cos(x) = 1/sec(x)

\[\implies \frac{\tan^2(x)}{\sec^2(x)} \color{red}{\implies \frac{\sec^2(x) - 1}{\sec^2(x)}} \; \; \; \: [Using \; \; Pythagorean \; \; Identity] \implies LHS\]

Answer 52

Either you prove your LHS equal to RHS or your RHS equal to LHS, Identity gets proved.. Do only one, both is not necessary, generally we prove LHS equal to RHS..

Answer 53

Oh i see

Answer 54

Sorry to bother too much....

Answer 55

It is okay.. :)