Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$ for the following given data

Let $A, B, C$ be real numbers such that

(i) $(\sin A, \cos B)$ lies on a unit circle centered at origin.

(ii) $\tan C$ and $\cot C$ are defined.

Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$.

My multiple attempts are as follows:-

Attempt $1$:

$$\sin^2A+\cos^2B=1$$ $$\tan^2C+\sin^2A-2\sin A\tan C+\cot^2C+\cos^2 B-2\cot C\cos B$$ $$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin C}{\cos C}+\dfrac{\cos C\cos B}{\sin C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos B}{\sin C\cos C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin^2C(\sin A-\cos B)+\cos B}{\sin C\cos C}\right)\tag{1}$$

Now from here how to proceed further.

Attempt $2$:

$$\sin^2A+\cos^2B=1$$ $$\sin^2A=\sin^2B$$ $$A=n\pi\pm B$$

Considering only the principal range, $A=B$, $A=-B$, $A=n\pi-B$, $A=n\pi+B$

Case $1$: $A=B,A=-B$

Put $B=A$ or $B=-A$ in equation $(1)$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos A}{\sin C\cos C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A+\alpha)}{\sin C\cos C}$$ $$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot \sin(A+\alpha)$$

So minimum value will be $3-2\sqrt{2}$

Case $1$: $A=n\pi-B,A=n\pi+B$

Put $B=n\pi-A$ or $B=A-n\pi$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C-\cos^2 C\cos A}{\sin C\cos C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A-\alpha)}{\sin C\cos C}$$

$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot\sin(A-\alpha)$$

So minimum value will be $3-2\sqrt{2}$

Any other way to solve this question?

Solution 1:

If $x=\tan C$, and we have $\cos^2 A=\cos ^2B$, we need to minimize $$(x-\sin A)^2+\left(\dfrac{1}{x}-\cos A\right)^2=x^2+\dfrac{1}{x^2}+1-2\left(x\sin A+\dfrac{\cos A}{x}\right)$$

So, we need to maximize $\left(x\sin A+\dfrac{\cos A}{x}\right)$.

It is known that maximum of $\alpha \sin \theta+\beta \cos \theta$ is $\sqrt{\alpha^2+\beta^2}$ , so it's maximum is $\sqrt{x^2+\dfrac{1}{x^2}}$, so our expression becomes $$x^2+\dfrac{1}{x^2}-2\sqrt{x^2+\dfrac{1}{x^2}}+1=\left(\sqrt{x^2+\dfrac{1}{x^2}}-1\right)^2$$

And clearly by A.M-G.M inequality, $x^2+\dfrac{1}{x^2}\geq 2$, so the minimum is $(\sqrt{2}-1)^2$.

Equality occurs at $x=1=\tan C$.

Solution 2:

Let $P = (\tan C, \cot C)$ lie on the curve $xy = 1$, $Q = (\sin A, \cos B)$ lie on the unit circle $x^2 + y^2 = 1$ and $O = (0, 0)$ be the origin.

It's easy to prove by AM-GM inequality that $PO \geqslant \sqrt2$, so $(\tan C - \sin A)^2 + (\cot C - \cos B)^2 = |PQ|^2 \geqslant (|PO| - |QO|)^2 \geqslant 3 - 2\sqrt 2$, where the equality is reached iff $P = (1, 1)$ and $Q = \left(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\right)$ or their images under O-reflection.