Question

tan(x+100°) = tanx (tan(x+50°))(tan(x-50°))

Find least positive value of x

Answer 1

30°

Answer 2

The given equation is: $\tan(x+100^\circ) = \tan x \tan(x+50^\circ) \tan(x-50^\circ)$

We can rewrite the equation as: $\frac{\tan(x+100^\circ)}{\tan x} = \tan(x+50^\circ) \tan(x-50^\circ)$

We use the identity $\tan(A+B)\tan(A-B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}$. Alternatively, and more usefully for this problem, we can use the product-to-sum identity for cosines: $\tan(A+B)\tan(A-B) = \frac{\sin(A+B)\sin(A-B)}{\cos(A+B)\cos(A-B)}$ Using $\sin(X)\sin(Y) = \frac{1}{2}[\cos(X-Y) - \cos(X+Y)]$ and $\cos(X)\cos(Y) = \frac{1}{2}[\cos(X-Y) + \cos(X+Y)]$: $\tan(A+B)\tan(A-B) = \frac{\cos((A+B)-(A-B)) - \cos((A+B)+(A-B))}{\cos((A+B)-(A-B)) + \cos((A+B)+(A-B))}$ $\tan(A+B)\tan(A-B) = \frac{\cos(2B) - \cos(2A)}{\cos(2B) + \cos(2A)}$

Let $A=x$ and $B=50^\circ$. Then $A+B = x+50^\circ$ and $A-B = x-50^\circ$. So, $\tan(x+50^\circ)\tan(x-50^\circ) = \frac{\cos(2 \times 50^\circ) - \cos(2x)}{\cos(2 \times 50^\circ) + \cos(2x)}$ $\tan(x+50^\circ)\tan(x-50^\circ) = \frac{\cos(100^\circ) - \cos(2x)}{\cos(100^\circ) + \cos(2x)}$

Substitute this back into the original equation: $\tan(x+100^\circ) = \tan x \left( \frac{\cos(100^\circ) - \cos(2x)}{\cos(100^\circ) + \cos(2x)} \right)$

Convert tangents to sines and cosines: $\frac{\sin(x+100^\circ)}{\cos(x+100^\circ)} = \frac{\sin x}{\cos x} \left( \frac{\cos(100^\circ) - \cos(2x)}{\cos(100^\circ) + \cos(2x)} \right)$

Cross-multiply: $\sin(x+100^\circ)\cos x (\cos(100^\circ) + \cos(2x)) = \cos(x+100^\circ)\sin x (\cos(100^\circ) - \cos(2x))$

Expand the terms: $\sin(x+100^\circ)\cos x \cos(100^\circ) + \sin(x+100^\circ)\cos x \cos(2x) = \cos(x+100^\circ)\sin x \cos(100^\circ) - \cos(x+100^\circ)\sin x \cos(2x)$

Rearrange terms to group common factors: $\cos(100^\circ) [\sin(x+100^\circ)\cos x - \cos(x+100^\circ)\sin x] = -\cos(2x) [\sin(x+100^\circ)\cos x + \cos(x+100^\circ)\sin x]$

Use the sum/difference formulas for sine: $\sin(A-B) = \sin A \cos B - \cos A \sin B$ $\sin(A+B) = \sin A \cos B + \cos A \sin B$

So, the equation becomes: $\cos(100^\circ) \sin((x+100^\circ)-x) = -\cos(2x) \sin((x+100^\circ)+x)$ $\cos(100^\circ) \sin(100^\circ) = -\cos(2x) \sin(2x+100^\circ)$

Use the identity $\sin(2A) = 2 \sin A \cos A$: $\frac{1}{2} \sin(2 \times 100^\circ) = -\cos(2x) \sin(2x+100^\circ)$ $\frac{1}{2} \sin(200^\circ) = -\cos(2x) \sin(2x+100^\circ)$

Since $\sin(200^\circ) = \sin(180^\circ+20^\circ) = -\sin(20^\circ)$: $\frac{1}{2} (-\sin(20^\circ)) = -\cos(2x) \sin(2x+100^\circ)$ $\frac{1}{2} \sin(20^\circ) = \cos(2x) \sin(2x+100^\circ)$

Use the product-to-sum identity for sine and cosine: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$. Let $A = 2x+100^\circ$ and $B=2x$. $\frac{1}{2} \sin(20^\circ) = \frac{1}{2} [\sin((2x+100^\circ)+2x) + \sin((2x+100^\circ)-2x)]$ $\frac{1}{2} \sin(20^\circ) = \frac{1}{2} [\sin(4x+100^\circ) + \sin(100^\circ)]$

Multiply by 2: $\sin(20^\circ) = \sin(4x+100^\circ) + \sin(100^\circ)$ $\sin(20^\circ) - \sin(100^\circ) = \sin(4x+100^\circ)$

Use the difference-to-product identity for sine: $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$. $2 \cos\left(\frac{20^\circ+100^\circ}{2}\right) \sin\left(\frac{20^\circ-100^\circ}{2}\right) = \sin(4x+100^\circ)$ $2 \cos(60^\circ) \sin(-40^\circ) = \sin(4x+100^\circ)$ $2 \left(\frac{1}{2}\right) (-\sin(40^\circ)) = \sin(4x+100^\circ)$ $-\sin(40^\circ) = \sin(4x+100^\circ)$

Since $-\sin \theta = \sin(-\theta)$: $\sin(-40^\circ) = \sin(4x+100^\circ)$

The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$, where $n$ is an integer. So, $4x+100^\circ = n \times 180^\circ + (-1)^n (-40^\circ)$.

We need to find the least positive value of $x$. Case 1: $n$ is an even integer, $n=2k$. $4x+100^\circ = 2k \times 180^\circ + (-1)^{2k} (-40^\circ)$ $4x+100^\circ = 360^\circ k - 40^\circ$ $4x = 360^\circ k - 140^\circ$ $x = 90^\circ k - 35^\circ$

For $k=0$, $x = -35^\circ$ (not positive). For $k=1$, $x = 90^\circ - 35^\circ = 55^\circ$. (This is a positive value) For $k=2$, $x = 180^\circ - 35^\circ = 145^\circ$.

Case 2: $n$ is an odd integer, $n=2k+1$. $4x+100^\circ = (2k+1) \times 180^\circ + (-1)^{2k+1} (-40^\circ)$ $4x+100^\circ = (2k+1) \times 180^\circ + 40^\circ$ $4x+100^\circ = 360^\circ k + 180^\circ + 40^\circ$ $4x+100^\circ = 360^\circ k + 220^\circ$ $4x = 360^\circ k + 120^\circ$ $x = 90^\circ k + 30^\circ$

For $k=0$, $x = 30^\circ$. (This is a positive value) For $k=1$, $x = 90^\circ + 30^\circ = 120^\circ$.

Comparing the positive values obtained: $55^\circ$ and $30^\circ$. The least positive value of $x$ is $30^\circ$.

We must also check for conditions where the tangents are undefined, i.e., when their arguments are $(m+1/2)\pi$. $x \neq (m+1/2)\pi$ $x+50^\circ \neq (m+1/2)\pi$ $x-50^\circ \neq (m+1/2)\pi$ $x+100^\circ \neq (m+1/2)\pi$

For $x=30^\circ$: $\tan(30^\circ)$ is defined. $\tan(30^\circ+50^\circ) = \tan(80^\circ)$ is defined. $\tan(30^\circ-50^\circ) = \tan(-20^\circ)$ is defined. $\tan(30^\circ+100^\circ) = \tan(130^\circ)$ is defined. None of these angles are odd multiples of $90^\circ$. So $x=30^\circ$ is a valid solution.

The final answer is $\boxed{\text{30°}}$.