Asked by Vijaya
Prove sqrt(Sec^2 A+Cosec^2 A)=TanA+CotA
Answers
Answered by
Mgraph
It is not true if (for example) A=3pi/4
Answered by
MathMate
Expand left side into sines and cosines:
sqrt(sec^2A+csc^2A)
=sqrt(1/cos^2A+1/sin^2A)
=sqrt((sin^2A+cos^2A)/(cos^2A sin^2A))
=sqrt(1/(cos^2A sin^2A))
=1/cosA sinA
Similarly expand right hand side:
tanA+cotA
=sinA/cosA + cosA/sinA
=(sin^2A + cos^2A)/(cosA sinA)
=1/(cosA sinA)
sqrt(sec^2A+csc^2A)
=sqrt(1/cos^2A+1/sin^2A)
=sqrt((sin^2A+cos^2A)/(cos^2A sin^2A))
=sqrt(1/(cos^2A sin^2A))
=1/cosA sinA
Similarly expand right hand side:
tanA+cotA
=sinA/cosA + cosA/sinA
=(sin^2A + cos^2A)/(cosA sinA)
=1/(cosA sinA)
Answered by
Mgraph
tan(3pi/4)=cot(3pi/4)=-1
sec^2(3pi/4)=csc^2(3pi/4)=2
sqrt(2+2)=-1-1 ?
sec^2(3pi/4)=csc^2(3pi/4)=2
sqrt(2+2)=-1-1 ?
Answered by
MathMate
Here square-root is taken of the square of the product of two functions, and not the numerical values.
To me it is justified to retain the signs of the original functions, namely sin(x) and cos(x) in the square-root.
So if we evaluate the functions after taking square-root,
LHS=1/(cos(3π/4)sin(3π/3)=-2
and
RHS=-2 as you have calculated.
As a compromise, we can say that the identity should read:
(Sec^2 A+Cosec^2 A)=(TanA+CotA)²
To me it is justified to retain the signs of the original functions, namely sin(x) and cos(x) in the square-root.
So if we evaluate the functions after taking square-root,
LHS=1/(cos(3π/4)sin(3π/3)=-2
and
RHS=-2 as you have calculated.
As a compromise, we can say that the identity should read:
(Sec^2 A+Cosec^2 A)=(TanA+CotA)²
Answered by
Mgraph
In the problem we must add
0<A<pi/2 that's all
0<A<pi/2 that's all
Answered by
MathMate
Yes, or in more general terms
kπ<A<kπ+π/2 k∈Z
kπ<A<kπ+π/2 k∈Z
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.