hlo mates plzz solve it thank you
Attachments:

Answers
Answered by
2
tanA=ntanB
=tanB=1/ntanA
=cotB=n/tanA
and, sinA=msinB
sinB=1/msinA
cosecBm/sinA
substituting the values of cot B and cosecB in cosec^2B-cot^2B=1,
m^2/sin^2A-n^2/tan^2A
m^/sin^2A-n^cos^2A/sin^2A=1
m^2-n^2cos^2A/sin^2A=1
m^2-n^2cos^2A=1
m^2-n^2cos^2A=1-cos^2A
m^2-1=n^2cos^2A-cos^2A
m^2-1=n^2-1×cos^2
m^2-1/n^2-1=cos^2
=tanB=1/ntanA
=cotB=n/tanA
and, sinA=msinB
sinB=1/msinA
cosecBm/sinA
substituting the values of cot B and cosecB in cosec^2B-cot^2B=1,
m^2/sin^2A-n^2/tan^2A
m^/sin^2A-n^cos^2A/sin^2A=1
m^2-n^2cos^2A/sin^2A=1
m^2-n^2cos^2A=1
m^2-n^2cos^2A=1-cos^2A
m^2-1=n^2cos^2A-cos^2A
m^2-1=n^2-1×cos^2
m^2-1/n^2-1=cos^2
Answered by
3
The solution is in the attachment....
Plz follow me.....☝☝☝☝
Attachments:

Similar questions