omicron
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I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]
The discussion revolves around solving two trigonometric problems: one involving the equation tan2A = cot40 and the other concerning cosec^2x = 1. The scope includes mathematical reasoning and problem-solving techniques related to trigonometric identities and functions.
Participants express varying levels of understanding and approaches to the problems, with some agreeing on the definitions and methods while others seek clarification. No consensus is reached on the solutions to the problems.
There are unresolved assumptions regarding the use of calculators and the interpretation of trigonometric identities. The discussion does not clarify all mathematical steps involved in reaching conclusions.
omicron said:I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]
Doing the question. I just don't know.Exactly what help do you need?
So are u saying that [tex]cosec^2x=1[/tex] can be also written as [tex]\frac{1}{sin^2x} = 1[/tex]As for the second problem, do you know that cosec x is defined as 1/sin x?
omicron said:Doing the question. I just don't know.
How do u find cot(40)? Don't u have to change it to [tex]\frac{1}{tan\theta}[/tex] or something like that?
So are u saying that [tex]cosec^2x=1[/tex] can be also written as [tex]\frac{1}{sin^2x} = 1[/tex]
omicron said:I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]
