cotangent(theta + 5)=tangent(theta +35)

Question

cotangent(Θ +5)=tangent(Θ + 35)

Answer 1

cot(x+5) = 1/tan(x+5) = tan(x+35)

tan(x+5)(tanx+35) = 1, use trig identities tan(a+b) = (tana +tanb)/(1-tanatanb)

= ((tanx + tan5)/(1-tanxtan5))((tanx +tan35)/(1-tanxtan35)^2)=1

= (tanx+tan5)(tanx+tan35)/(1-tanxtan5)^2 =1

= (tan^2(x)+tanxtan35+tan5tanx + tan5tan35)/(1-2tanxtan35 + tan^2(x)tan^2(35))= 1

tan^2(x) + tanxtan35 +tan5tanx + tan5tan35 = 1 - 2tanxtan35 + tan^2(x)tan^2(35)

tan^2(x) +3tanxtan35 + tan5tanx +tan5tan35 - tan^2(x)tan^2(35) = 1 let y = tanx

(1-tan^2(35)y^2 +(3tan35 +tan5)y +tan5tan35 -1 = 0 use the quadratic formula

y = (-3tan35- tan5)/2(1-tan^2(35) +/-(1/2(1-tan^2(35))sqr((3tan35+tan5)^2 -4(1-tan^2(35)(tan5tan35 -1) = tanx

x = tan-1(y)

or graph and find x intercepts

x = about 1.5 or 2.75 + pin where n = any integer and pi = about 3.141593 radians

or

calculate tan 5=-3.38 and tan 35 =.47 radians

use that to simplify 1st method above

might also check whether it's a possible identity and plug in some angle for x such as zero

cot(x+5)= tan(x+35)

but cot0 doesn't =tan0,

so it's an equation valid only for certain values of x