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?
Best Answer
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 $x=1=\tan C$.