Trig
posted by VD .
(csc(B)+cot (B))(csc(B)cot(B))

You have the pattern of "difference of squares"
(csc(B)+cot (B))(csc(B)cot(B))
= csc^2 B  cot^2 B
But csc^ B = 1 + cot^2B , one of the identies from
sin^2 B + cos^2 B= 1
so csc^2 B  cot^2 B
= 1 + cot^2 B  cot^2 B = 1
Respond to this Question
Similar Questions

Calculus
Hello, I'm having trouble with this exercise. Can you help me? 
Trigonometry
Hello all, In our math class, we are practicing the trigonometric identities (i.e., sin^2(x)+cos^2(x)=1 or cot(x)=cos(x)/sin(x). Now, we are working on proofs that two sides of an equation are equal (for example, sin(x)*csc(x)=1; sin(x)csc(x)=sin(x)/sin(x)=1; … 
Math  Trig
I'm trying to verify these trigonometric identities. 1. 1 / [sec(x) * tan(x)] = csc(x)  sin(x) 2. csc(x)  sin(x) = cos(x) * cot(x) 3. 1/tan(x) + 1/cot(x) = tan(x) + cot(x) 4. csc(x)/sec(x) = cot(x) 
trig 30
For csc^2 A1/cot A csc A, what is the simplest equivalent trig expression? 
trig
prove the identity: csc x1/csc x+1 = cot^x/csc^x+2 csc x+1 
trig
verify : [sec(x) / csc(x)  cot(x)]  [sec(x) / csc(x) + cot(x)] = 2csc(x) 
trigonometry repost
Reduce (csc^2 x  sec^2 X) to an expression containing only tan x. (is this correct? 
pre cal
can someone please show me how to verify this identity? 
Trig verifying identities
I am having trouble with this problem. sec^2(pi/2x)1= cot ^2x I got : By cofunction identity sec(90 degrees  x) = csc x secx csc1 = cot^2x Then split sec x and csc1 into two fractions and multiplied both numerator and denominators … 
trig
verify (csc^41)/cot^2x=2+cot^2x So this is what I have so far on the left side (csc^2x+1)(cscx+1)(cscx1)/cot^2x =(csc^2x+1)(cot^2x)/cot^2x i think I'm doing something wrong. Please help!