Prove sqrt(Sec^2 A+Cosec^2 A)=TanA+CotA
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)
tan(3pi/4)=cot(3pi/4)=-1
sec^2(3pi/4)=csc^2(3pi/4)=2
sqrt(2+2)=-1-1 ?
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 prove the equation sqrt(Sec^2 A + Cosec^2 A) = Tan A + Cot A, we'll need to start by simplifying each side of the equation individually and then show that they are equal.
Let's begin with the left side of the equation:
sqrt(Sec^2 A + Cosec^2 A)
Recall the trigonometric identities:
Sec^2 A = 1/(Cos^2 A)
Cosec^2 A = 1/(Sin^2 A)
Substituting these identities into the equation, we have:
sqrt(1/(Cos^2 A) + 1/(Sin^2 A))
To add these fractions, we need to find a common denominator. The common denominator here is (Cos^2 A)*(Sin^2 A). Thus, we can rewrite the equation as:
sqrt((Sin^2 A + Cos^2 A)/(Cos^2 A * Sin^2 A))
Now, notice that the numerator Sin^2 A + Cos^2 A is equal to 1 (by the Pythagorean identity). Substituting this into the equation, we get:
sqrt(1/(Cos^2 A * Sin^2 A))
Next, we can simplify the expression inside the square root:
sqrt(1/[(Cos A * Sin A)^2])
Taking the square root of the denominator, we have:
1/(Cos A * Sin A)
Now, let's simplify the right side of the equation:
Tan A + Cot A
Using the trigonometric identities, we can rewrite Tan A and Cot A:
Sin A / Cos A + Cos A / Sin A
To add these fractions, we need a common denominator, which is (Sin A) * (Cos A). Rewriting the equation with this common denominator, we have:
(Sin^2 A + Cos^2 A) / (Cos A * Sin A)
Again, the numerator Sin^2 A + Cos^2 A is equal to 1. Substituting this, we get:
1 / (Cos A * Sin A)
Comparing the simplified left side and right side of the equation, we can see that they are equal:
1/(Cos A * Sin A) = 1/(Cos A * Sin A)
Therefore, we have proved that sqrt(Sec^2 A + Cosec^2 A) = Tan A + Cot A.
In the problem we must add
0<A<pi/2 that's all
Yes, or in more general terms
kπ<A<kπ+π/2 k∈Z