Question

Prove

Answer 1

\[\sqrt{\sec^2A+cosec^2A}=tanA+cotA\]

Answer 2

use identitites for sec^2 and cosec^2

Answer 3

then write 2 = 2*tan * cot

Answer 4

how will the root remove

Answer 5

it will , trust me...use them

Answer 6

ok wait

Answer 7

You guys gonna duke it out?

Answer 8

what u got after using the identities ?

Answer 9

sqrt(2-tan^A-cot^2A)

Answer 10

*tan^2A

Answer 11

- *minus* or + ?

Answer 12

check the identities again!

Answer 13

sec^A = 1-tan^A
cosec^A = 1-cot^2A

Answer 14

Sorry, but can't you just convert it to sines and cosines and brute force it?

Answer 15

sorry I am mistyping a lot

Answer 16

sec^A = 1+tan^2A
cosec^A = 1+cot^2A

Answer 17

wio, this way its easier.

Answer 18

cot^2A+cosec^2A = 1 ??

Answer 19

\(\sqrt{2+\tan^2x+\cot^2x}=\sqrt{2\tan x\cot x+\tan^2x+\cot^2x}\)
now do you see a form a^2+2ab+b^2 under the root ??

Answer 20

noooooooooooooooo.............
cot^2A+1=cosec^2A

Answer 21

Well there is always...\[
\sqrt{\frac{1}{\cos^2A}+\frac{1}{\sin^2A}} = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}
\]

Answer 22

ofcourse! i never said you can't
i am just taking shorter way out...

Answer 23

i forgot that cot^2A -cosec^2A = 1

Answer 24

I know but \(2 = 2\tan A\cot A\)? Pretty subtle.

Answer 25

cosec^2 A - cot^2 A =1

Answer 26

what happened in the second step u typed above

Answer 27

after sqrt(2+tan^2x+cot^2x)

Answer 28

i wrote 2=2tan*cot
so that u get the form a^2+2ab+b^2
if you are not comfortable with this, convert everything into sin and cos

Answer 29

can we multiply and divide by 2tanAcotA

Answer 30

why would you do that ?

Answer 31

\(\sqrt{2+\tan^2x+\cot^2x}=\sqrt{2\tan x\cot x+\tan^2x+\cot^2x} \\ = (\sqrt {\tan x+\cot x})^2\)

Answer 32

there must be a way to get tanAcotA with doing this 2=2tanAcotA

Answer 33

*without

Answer 34

Add and subtract ??

Answer 35

convert everything to sin and cos, as wio said

Answer 36

First you should know this:\[\tan^2x+1=\sec^2x\]\[\cot^2x+1=\csc^2x\]Squaring both sides of the equation gives:\[\sec^2A + \csc^2A=\tan^2A+2 \tan A \cot A+\cot^2A \]\[ \sec^2A + \csc^2A = \tan^2A + \cot^2A + 2\]Using the two identities stated in the beginning we get:\[(\tan^2A+1) + (\cot^2A+1)=\tan^2A+\cot^2A+2 \rightarrow R.S = L.S\]Therefore, right side = left side.

@koli123able

Answer 37

why to square both sides of given equation, when we can prove without doing it ? O.o

Answer 38

It's easier and more obvious.

Answer 39

Get it? @koli123able

Answer 40

i would never change the statement given us to be proved -_-

Answer 41

i always prove using a single side

Answer 42

Just because you do what you always do doesn't mean you can't do something else.

Answer 43

He @genius12 Did you know \(1=-1\)
\[
(1)^2 = (-1)^2 \\
1=1
\]

Answer 44

^that

Answer 45

And what does that have to do with anything?

Answer 46

I squared both sides.

Answer 47

oh, you didn't get it ?
squaring both sides, attaches an extraneous root.

Answer 48

I assumed what I was proving was true... turns out I'm right.

Answer 49

I dont think you can do that, wio.You should only work with one side and then end up with a value that equals the other side.

Answer 50

^exactly what he meant.

Answer 51

Sometimes I feel that people are very inclined to only doing what they've been taught and thinking that anything else apart from that is wrong.

For example, to prove the equation, we don't have to start from one side and prove it to the rest like we always do. I can change one side to an equivalent expression and then solve the other side to equal to that. It's still the same thing, but a little apart from the traditional method of simply taking one side and proving it to be the other side.

Answer 52

\(\sqrt a=x \)
and
\(a=x^2\)
are different.
first has only one answer for 'x'
2nd is quadratic and has 2 values of 'x'

Answer 53

thats the point here!
the equation will NOT be EQUIVALENT once you square both sides.

Answer 54

so, not same thing...

Answer 55

I know what you're saying. Relax. And I'm keeping everything in mind. It's 5:10 AM here. I am a little tired. lol.

Answer 56

^good one!!! (y) @hartnn

Answer 57

i am relaxed...sorry if you thought i was shouting...i was just emphasizing...

Answer 58

2=2tanAcotA ??? <---- there must be a way else this way

Answer 59

K, let's change the tanA + cotA to 1/cosAsinA. Then simply change the stuff under the square root to 1/cos^2A + 1/sin^2A. After you simply and take the square root, you get the same thing as the right side.

Happy @hartnn?

Answer 60

i am trying and i will come up with that way!! for sure

Answer 61

just my thoughts...
we always readily think of replacing \(\sin^2x+\cos^2x\) with 1.
but why not think of replacing 1 by \(\sin^2x+\cos^2x\) ,
if things get much simplified by that!
same here for tan x cot x =1
this thought process will help you so much in integration....

Answer 62

what should i replace by 1? tanx cot x??

Answer 63

in your question, you got 2+....
i replaced 2 by 2tan x cot x
2=2*1 = 2*(tan x cot x)
didn't that directly gave us the form a^2+2ab+b^2
and became much simpler to solve...

Answer 64

that step is awesome i love it !! @hartnn

Answer 65

and thanks

Answer 66

welcome ^_^

Answer 67

By the way @hartnn What about the absolute value?

Answer 68

\[
\sqrt{(a+b)^2} = |a+b|
\]

Answer 69

yes, thats true
\(\sqrt{\sec^2A+cosec^2A}=|\tan A+\cot A|\)
the question itself ignored it!

Answer 70

No...
\[\sqrt{a^2} = a\]
But
\[x^2=a\]\[x=\pm a\]

Answer 71

\[
\sqrt{a^2} = |a|
\]Is practically a definition of the absolute value.

Answer 72

\[x^2=a^2\]\[x=\pm a\]

Answer 73

\(\sqrt{a^2}=|a|\)
if you just write 'a' , what about negative value of a ?

Answer 74

\[
\sqrt{(-1)^2} =\sqrt{1}=1
\]Tell me where I went wrong? @Callisto

Answer 75

Or do you also think 1=-1? =)

Answer 76

Hey @koli123able
See if you can use \(3\pi/4\) to disprove it by counter example

=]

Answer 77

My mistake, sorry!