Solve the equation (tan theta -1) ÷ (tan theta +1) = tan theta , for 0⁰ =< theta =< 180⁰?

(tanθ - 1) ÷ (tanθ + 1) = tanθ
[(tanθ - 1) ÷ (tanθ + 1)] (tanθ + 1) = tanθ (tanθ + 1)
tanθ - 1 = tan² θ + tanθ
tan²θ = -1
tanθ has no real solution
Hence, θ has no real solution.

We have (tanθ - 1)/(tanθ + 1) = tanθ   

so, tanθ - 1 = tanθ(tanθ + 1)

i.e. tanθ - 1 = tan²θ + tanθ

Hence,  tan²θ = -1....no solution

:)> 

Let alpha=A.
(tanA-1)/(tanA+1)=tanA, 0*=<A<=180*
=>
(sinA-cosA)/(sinA+cosA)=sinA/cosA
[assume cosA=/=0 or A=/=90*]
=>
sinAcosA-cos^2(A)=sin^2(A)+sinAcosA
=>
sin^2(A)+cos^2(A)=0
=>
1=0
is a contradiction, No solution.

(tanx -1)/(tanx +1) = tanx, x in (0,180). Put (s,c,t) = (sinx,cosx,tanx).;
Then (t-1)/(t+1) = t, ie., t-1 = t(t+1), ie., t-1 = t^2+t, ie., t^2 = -1, which is absurd. Problem has no
solution.

(tan(t) - 1) / (tan(t) + 1) = tan(t)
tan(t) - 1 = tan(t) * (tan(t) + 1)
tan(t) - 1 = tan(t)^2 + tan(t)
-1 = tan(t)^2
No solution

Recall the identity:

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = θ and suppose that: b = 1

sin(θ + 1) = sin(θ).cos(1) + cos(θ).sin(1) ← memorize this result as (1)


Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = θ and suppose that: b = 1

cos(θ + 1) = cos(θ).cos(1) - sin(θ).sin(1) ← memorize this result as (2)


Recall the identity:

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → suppose that: a = θ and suppose that: b = 1

sin(θ - 1) = sin(θ).cos(1) - cos(θ).sin(1) ← memorize this result as (3)


Do you know this identity?

cos(a - b) = cos(a).cos(b) + sin(a).sin(b) → suppose that: a = θ and suppose that: b = 1

cos(θ - 1) = cos(θ).cos(1) + sin(θ).sin(1) ← memorize this result as (4)



tan(θ - 1) = sin(θ - 1)/cos(θ - 1)

tan(θ - 1) = (3) / (4)

tan(θ - 1) = [sin(θ).cos(1) - cos(θ).sin(1)] / [cos(θ).cos(1) + sin(θ).sin(1)]

tan(θ - 1) = [sin(θ).cos(1) - cos(θ).sin(1)] / [cos(θ).cos(1) + sin(θ).sin(1)] → you divide by cos(1)

tan(θ - 1) = [sin(θ) - cos(θ).tan(1)] / [cos(θ) + sin(θ).tan(1)] → you divide by cos(θ)

tan(θ - 1) = [tan(θ) - tan(1)] / [1 + tan(θ).tan(1)]


tan(θ + 1) = sin(θ + 1)/cos(θ + 1)

tan(θ + 1) = (1) / (2)

tan(θ + 1) = [sin(θ).cos(1) + cos(θ).sin(1)] / [cos(θ).cos(1) - sin(θ).sin(1)] → you divide by cos(1)

tan(θ + 1) = [sin(θ) + cos(θ).tan(1)] / [cos(θ) - sin(θ).tan(1)] → you divide by cos(θ)

tan(θ + 1) = [tan(θ) + tan(1)] / [1 - tan(θ).tan(1)]


= tan(θ - 1) + tan(θ + 1)

= { [tan(θ) - tan(1)] / [1 + tan(θ).tan(1)] } + { [tan(θ) + tan(1)] / [1 - tan(θ).tan(1)] }

= { [tan(θ) - tan(1)].[1 - tan(θ).tan(1)] + [tan(θ) + tan(1)].[1 + tan(θ).tan(1)] } / { [1 + tan(θ).tan(1)].[1 - tan(θ).tan(1)] }

= { tan(θ) - tan²(θ).tan(1) - tan(1) + tan(θ).tan²(1) + tan(θ) + tan²(θ).tan(1) + tan(1) + tan(θ).tan²(1) } / { [1 + tan(θ).tan(1)].[1 - tan(θ).tan(1)] }

= { 2.tan(θ) + 2.tan(θ).tan²(1) } / { [1 + tan(θ).tan(1)].[1 - tan(θ).tan(1)] }

= 2.tan(θ).[1 + tan²(1)] / { [1 + tan(θ).tan(1)].[1 - tan(θ).tan(1)] }

= 2.tan(θ).[1 + tan²(1)] / { 1 - tan(θ).tan(1) + tan(θ).tan(1) - tan²(θ).tan²(1) }

= 2.tan(θ).[1 + tan²(1)] / [1 - tan²(θ).tan²(1)]


Your equation

tan(θ - 1) + tan(θ + 1) = tan(θ)

2.tan(θ).[1 + tan²(1)] / [1 - tan²(θ).tan²(1)] = tan(θ)

2.tan(θ).[1 + tan²(1)] = tan(θ).[1 - tan²(θ).tan²(1)]

2.tan(θ).[1 + tan²(1)] - tan(θ).[1 - tan²(θ).tan²(1)] = 0

tan(θ) * { 2.[1 + tan²(1)] - [1 - tan²(θ).tan²(1)] } = 0

tan(θ) * [2 + 2.tan²(1) - 1 + tan²(θ).tan²(1)] = 0

tan(θ) * [1 + 2.tan²(1) + tan²(θ).tan²(1)] = 0


First case:

tan(θ) = 0

θ = 0

θ = 180


Second case:

1 + 2.tan²(1) + tan²(θ).tan²(1) = 0

tan²(θ).tan²(1) = - 1 - 2.tan²(1)

tan²(θ) = [- 1 - 2.tan²(1)] / tan²(1)

tan²(θ) = - [1 + 2.tan²(1)] / tan²(1) ← no possible because a square cannot be negative

→ Solution = { 0 ; 180 }