Verify the identity - Pre-calculus

Question

Verify the Identity

 

cot²(x) − tan²(x) = csc²(x) − sec²(x)

Answer 1

Rewrite the LEFT side of this identity by replacing cot x with cos x / sin x and replacing tan x by sin x / cos x.

 

Then find a common denominator for this left side (it will be product of squares of sin x and cos x); this gives a numerator of cos⁴x - sin⁴x.  You factor this difference of squares into (cos²x - sin²x)(cos²x + sin²x) but you can replace that sum-of-squares factor by 1.

 

Your left side is now (cos²x - sin²x) divided by (cos²x sin²x).  Now when you divide this common denominator into each of the numerator's terms, you get 1/sin²x  -  1/cos²x, and this is the entire left side, which is equivalent to the original right side since reciprocal of sine is cosecant, reciprocal of cosine is secant.

 

Q.E.D.

Answer 2

Hi Sydney,

You are given an identity. Let us designate the right-hand side as RHS, and work on it to arrive at what the left-hand side (LHS) is.

cot²(x) − tan²(x) = csc²(x) − sec²(x)   -----(1)

 

 

RHS = csc²(x) − sec²(x)                     -----(2)

From definition

sin²(x) + cos² (x) = 1                         -----(3),  Divide everything on left and right by sin²(x)

1  + cos²(x)/sin²(x) = 1/sin²(x)      [cos(x)/sin(x) = cot(x) and 1/sin(x) = csc(x)]

1 + cot²(x) = csc²(x)                          -----(4) plug this in

 

csc(x) = 1/sin(x) and sec(x)= 1/cos(x) plug them in (2)

RHS= 1 + cot²(x) - sec²(x)                 -----(5)  Now divide (3) by cos²(x)

sin²(x)/cos²(x) + 1 = 1/cos²(x)       [sin(x)/cos(x) = tan(x), 1/cos(x) = sec(x)]

tan²(x) + 1 = sec²(x)                        ------(6) Plug this into (5)

RHS = 1 + cot²(x) - (tan²(x) +1)

       =1 + cot²(x) - tan²(x) - 1

       = cot²(x) - tan²(x)

LHS = cot²(x) - tan²(x)

LHS = RHS ----qed