Question

Let A={A:sinA =tanA} and
B={A:cosA=1} be two sets.Then:
A is a subset of B ...How??

Answer 1

My ans is B is a subset of A

Answer 2

I got these solution sets:
\[\large \begin{gathered}
A = \left\{ {x \in \mathbb{R}|x = k\pi ,\quad k \in \mathbb{Z}} \right\} \cup \left\{ {x \in \mathbb{R}|x = 2h\pi ,\quad h \in \mathbb{Z}} \right\} - \hfill \\
- \left\{ {x \in \mathbb{R}|x = \frac{\pi }{2} + q\pi ,\quad q \in \mathbb{Z}} \right\} \hfill \\
\hfill \\
B = \left\{ {x \in \mathbb{R}|x = 2k\pi ,\quad k \in \mathbb{Z}} \right\} \hfill \\
\end{gathered} \]

Answer 3

That clearly means the same as i came up with ...
Ur ans is also B is a subset of A

Answer 4

yes!

Answer 5

Ans key says that A is a subset of B

Answer 6

sincerely I don't know what to say

Answer 7

Ans key is wrong perhaps!

Answer 8

please note that:
the solutions we have this:
\[\begin{gathered}
\sin \theta = \tan \theta \Rightarrow \hfill \\
\hfill \\
\Rightarrow \sin \theta \left( {1 - \frac{1}{{\cos \theta }}} \right) = 0 \Rightarrow \hfill \\
\hfill \\
\Rightarrow \sin \theta \left( {\frac{{\cos \theta - 1}}{{\cos \theta }}} \right) = 0 \Rightarrow \hfill \\
\hfill \\
\Rightarrow \sin \theta = 0 \cup \cos \theta - 1 = 0 \hfill \\
\end{gathered} \]
for first equation

Answer 9

oops.. I meant ...the solutions we have are these...

Answer 10

whereas for second equation are these ones:
\[\cos \theta - 1 = 0\]

Answer 11

Exactly i too solve in this pattern only

Answer 12

Solved *

Answer 13

That clearly means that A can be 0,π,2π,....nπ for set A
Whereas for set B we just have A=0,2π,4π,....,2nπ

Answer 14

I remembered when I learned for first time the measure theory, so
in the topological sense, we can say that set A is more fine than the set B

Answer 15

nevertheless I don't think that it can be the needed explanation

Answer 16

So it is clear than B has got to be the subset of A and not the reverse of it

Answer 17

in the set theory I think so!

Answer 18

please, ask to another helper too

Answer 19

Who??

Answer 20

I think that we can try to get help from @ganeshie8

Answer 21

also @mayankdevnani can help us, I think!

Answer 22

I think it is sufficient to prove below :
\[\cos A = 1~~\implies~~ \sin A = \tan A \]

Answer 23

Yeah! That again means the same...

Answer 24

Is that hard to prove ?

Answer 25

but converse is also true !

Answer 26

converse isn't true

Answer 27

I didn't get you @ganeshie8 ..
Are u agreeing with ans that A is a subset of B ?

Answer 28

Answer is correct !

Answer 29

A is a subset of B

Answer 30

i have proved it !

Answer 31

How can we say than A is a subset of B when it is very clear that A contains more no . of elements than that in B?

Answer 32

Wait,lemme post the solution

Answer 33

Set A :-
Equate the general solutions:-
\[\large \bf n \pi +(-1)^n \alpha=n \pi+\alpha\]
\[\large \bf (-1)^n \alpha-\alpha=0\]
\[\large \bf \alpha[(-1)^n-1]=0\]
\[\large \bf \alpha=0~or~(-1)^n=1\]
where, n=0,2,4,6......


Set B :-
Equate the general solution :-
\[\large \bf cosA=1=\cos0\]
\[\large \bf cosA~where~A=2n \pi \pm \alpha\]
\[\large \bf A=2n \pi ~and~\alpha=0\]
where, n=0,1,2,3,4,..........

Answer 34

We can clearly see that A is a subset of B by noticing the `n`

Answer 35

@ganeshie8 @Michele_Laino i think it might helps !

Answer 36

This method works only when we are given like sinC =sinD then only can we write as
C=nπ+(-1)^n (D)
N furthermore why can't we do it as suggested by @michele_laino
It was the appropriate method too...

Answer 37

If u do it in that manner u will also come with the same ans as he and i got

Answer 38

What !

Answer 39

we have given sinA=tanA

Answer 40

in LHS, i wrote the general solution of sin
in RHS, i wrote the general solution of tan

Answer 41

Yep that doesn't mean we can equate general solutions of the two ....

Answer 42

we can !

Answer 43

can you give me a good reason why we can't ?

Answer 44

I don't get why you guys are trying to solve the given equations

Answer 45

@ganeshie8 am i wrong in my method ?

Answer 46

One sec...
Wouldn't you agree that proving `A is a subset of B` is same as proving the below statement :
\[\cos A = 1~~\implies~~ \sin A = \tan A\]

?

Answer 47

It's the definition of subset.\[
A\subseteq B \equiv \forall x \quad x\in A\implies x \in B
\]

Answer 48

No, i think this implies B is a subset of A ...

Answer 49

The equation ends up being: \[
\sin A = \tan A \implies \cos A = 1
\]But this is false, particularly when \(\sin A=0\).

Answer 50

I feel skughtly dyslexic at the moment, but do you see we don't have to solve anything here ?

Answer 51

Sir bt we can say that since set A has solutions {0,π,2π,3π,4π,.....}
And set B has solutions {0,2π,4π,....}
Which implies clearly that set B has got to be the subset of A and not the reverse of it...

Answer 52

Suppose \(A=\pi\). Then \(\sin A = 0 = \tan A\), so it is in \(A\); hoever \(\cos A = -1 \neq 1\) so it is not in \(B\).

Answer 53

Exactly @wio

Answer 54

sin A = tan A
A = n pi
A: 0,pi, 2pi, 3pi, ....

cos A =1
A =2n pi
B: 0, 2pi, 4pi, ...

B is the subset of A

Answer 55

Good god! You agreed with my ans @hartnn

Answer 56

So it's set now that the ans key is wrong ...bcoz the final ans is this only :
B is a subset of A ....
Thanks guys...

Answer 57

i bet your answer is wrong !

Answer 58

Looks good to me

Answer 59

mayank, you're saying 'n' CANNOT be = 1 here :
\(\large \bf n \pi +(-1)^n \alpha=n \pi+\alpha\)

but when \(\alpha = 0\), 'n' can indeed be = 1.
note : you ALSO have a "OR" in your answer.

---> when you equate the general solutions, you only take those which are common to both functions!

Answer 60

hmm

Answer 61

from \(\sin \theta = 0 \cup \cos \theta - 1 = 0\) also,
it is very clear that set A has solutions for \(\sin \theta = 0\) more than that of set B.

Answer 62

thank you !

Answer 63

well i applied that concept first
and when i red the answer is opposite, i changed my mind
lol

Answer 64

it is a good habit to challenge stuff mayank!
either you learn new things or you teach things that are new to others :)

keep it up! :)