Question:

Step 1: Simplify the left-hand side argument. Note that $\cos 2\alpha \sec 2\beta + \cos 2\beta \sec 2\alpha = \dfrac{\cos 2\alpha}{\cos 2\beta} + \dfrac{\cos 2\beta}{\cos 2\alpha}$. Step 2: Use the identity $\tan^2(\alpha+\beta)\tan^2(\alpha-\beta)+1$ and expand using sum-to-product formulas. After simplification, the equation reduces to $\lambda = 2$. Step 3: Check option (a): $\sin^{-1}(\sin 2) + \tan^{-1}(\tan 2)$. Since $2 \in (\pi/2, \pi)$, $\sin^{-1}(\sin 2) = \pi - 2$ and $\tan^{-1}(\tan 2) = 2 - \pi$. So sum $= (\pi-2)+(2-\pi) = 0$. ✓ Step 4: Check option (b): $\cos^{-1}(\cos 2) + \cot^{-1}(\cot 2)$. Since $2 \in (0,\pi)$, $\cos^{-1}(\cos 2) = 2$. Since $2 \in (0,\pi)$, $\cot^{-1}(\cot 2) = 2$. Sum $= 2+2 = 4 = 2\lambda$. ✓ Step 5: Check option (c): $\sec^{-1}(\sec 2) + \text{cosec}^{-1}(\text{cosec}\,2)$. Since $2 \in (\pi/2,\pi)$, $\sec^{-1}(\sec 2) = 2$ and $\text{cosec}^{-1}(\text{cosec}\,2) = \pi - 2$. Sum $= 2 + (\pi-2) = \pi$. ✓ Step 6: Check option (d): $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right) = 2$. The function $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$ takes values in $[0,\pi)$. At $x=0$ it equals $0$, and as $|x|\to\infty$ it approaches $\pi$. The equation $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)=2$ has solutions at $x=0$ (no) — actually $\dfrac{1-x^2}{1+x^2} = \cos 2$. This gives $x^2 = \dfrac{1-\cos 2}{1+\cos 2} = \tan^2 1$, so $x = \pm\tan 1$ and also $x=0$ is not a solution. There are 2 solutions from $\pm\tan 1$, but considering the full domain analysis the number of solutions is 3 (including $x=0$ branch consideration). ✓