Answer
Verify the following identity:
$(tan(x) + cot(x))^4 = sec(x)^4 csc(x)^4$
Work Step by Step
Verify the following identity:
$(tan(x) + cot(x))^4 = sec(x)^4 csc(x)^4$
Write cotangent as cosine/sine, cosecant as 1/sine, secant as 1/cosine and tangent as sine/cosine:
$((cos(x))/(sin(x)) + (sin(x))/(cos(x)))^4 = ^?( 1/(cos(x)) ^4 ) ( 1/(sin(x)) ^4 )$
$(1/(sin(x)))^4 (1/(cos(x)))^4 = 1/(cos(x)^4 sin(x)^4)$:
$((cos(x))/(sin(x)) + (sin(x))/(cos(x)))^4 = ^?1/(cos(x)^4 sin(x)^4)$
Put $(cos(x))/(sin(x)) + (sin(x))/(cos(x))$ over the common denominator $sin(x) cos(x)$: $(cos(x))/(sin(x)) + (sin(x))/(cos(x)) = (cos(x)^2 + sin(x)^2)/(cos(x) sin(x))$:
$(cos(x)^2 + sin(x)^2)/(cos(x) sin(x))^4 = ^?1/(cos(x)^4 sin(x)^4)$
Multiply each exponent in $(cos(x)^2 + sin(x)^2)/(sin(x) cos(x))$ by $4$:
$(cos(x)^2 + sin(x)^2)^4/(cos(x)^4 sin(x)^4) = ^?1/(cos(x)^4 sin(x)^4)$
Multiply both sides by $sin(x)^4 cos(x)^4$:
$(cos(x)^2 + sin(x)^2)^4 = ^?1$
$sin(x)^2 = 1 - cos(x)^2$:
$(cos(x)^2 + 1 - cos(x)^2)^4 = ^?1$
$cos(x)^2 + 1 - cos(x)^2 = 1$:
$1^4 = ^?1$
$1^4 = 1$:
$1 = ^?1$
The left hand side and right hand side are identical, thus verifying the identity.
