Question

eliminate θ between cosecθ-sinθ=m and secθ-cosθ=n

Answer 1

\[cosec\ \theta -\sin \theta=m\]
@samigupta8 Please tell me what's the relation between sin and cosec?

Answer 2

don't u know it

Answer 3

I do know, but I'm asking you so that you also take part in the process of answering

Answer 4

cotθcosθ=m

Answer 5

Let me rephrase, how cosec x and sin x are related?

Answer 6

yes

Answer 7

1/sinθ-sinθ=m

Answer 8

yes, so we'll get
\[\frac{1-\sin^2 \theta}{\sin \theta}=m\]
or
\[\frac{\cos^2 \theta}{\sin \theta} =m\]
or
\[\cot \theta \cos \theta =m\]Can you simplify the second equation?

Answer 9

\[\sec \theta -\cos \theta =n\]

Answer 10

sinθtanθ=n

Answer 11

then next

Answer 12

Yes, I'm thinking

Answer 13

kk.....

Answer 14

OK, let's multiply these two, we'd get
\[\sin \theta \cos \theta= mn\]
Now we'll square and add the two equations
\[(cosec \theta -\sin \theta)^2+(\sec \theta -\cos \theta)^2=m^2+n^2\]
\[cosec^2 \ \theta +\sin^2 \theta-1+\sec^2 \theta +\cos^2 \theta -1=m^2+n^2\]
\[cosec^2 \ \theta+\sec^2 \theta +1-1-1=m^2+n^2\]
Do you understand this?

Answer 15

ya...

Answer 16

Next we'll write cosec in terms of sin and sec in terms of cos. Can you try?
\[\cos \theta\sin \theta =mn\]

Answer 17

i m getting it as 1-m^2 n^2=(m^2-n^2)m^2 n^2

Answer 18

Check again, I got
\[(m^2+n^2+1)(mn)^2=1\]

Answer 19

i got it as (mn)^2+(mn)^2(m^2+n^2=1

Answer 20

yeasss... u are ryt

Answer 21

Interim steps are
\[\frac{1}{\sin^2 x}+\frac{1}{\cos^2 x}-1=m^2+n^2\]
\[\frac{\cos^2 x+\sin^2 x}{\cos^2 x \times \sin^2 x}-1=m^2+n^2\]
\[\frac{1}{(mn)^2}-1=m^2+n^2\]

Answer 22

btt answer is given as(m^2n)^2/3+(mn^2)^2/3=1

Answer 23

how is this ans coming