The number of solutions of tan⁻¹4x + tan⁻¹6x = π/6, where -1/(2√6) < x < 1/(2√6), is equal to

For inverse tangent, the addition formula is $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ when $ab < 1$. This is critical condition - if $ab = 1$, the sum is undefined as denominator becomes zero and argument goes to infinity meaning angle is $\pm\frac{\pi}{2}$. If $ab > 1$, we must add or subtract $\pi$ based on signs: when both $a,b > 0$ and $ab > 1$, add $\pi$; when both $a,b < 0$ and $ab > 1$, subtract $\pi$. The formula comes from tangent addition: $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$. Setting $\alpha = \tan^{-1}a$ and $\beta = \tan^{-1}b$ gives the result. For inverse sine and cosine: $\sin^{-1}a + \sin^{-1}b = \sin^{-1}(a\sqrt{1-b^2} + b\sqrt{1-a^2})$ when result is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Similarly $\cos^{-1}a + \cos^{-1}b = \cos^{-1}(ab - \sqrt{(1-a^2)(1-b^2)})$ when result is in $[0,\pi]$. These formulas are powerful for solving inverse trig equations and appear frequently in JEE Main and Advanced. Always verify the condition before applying formula.

Understanding domain and range is crucial for inverse trig. For $\sin^{-1}x$: domain is $[-1,1]$, range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For $\cos^{-1}x$: domain is $[-1,1]$, range is $[0,\pi]$. For $\tan^{-1}x$: domain is all real numbers $(-\infty, \infty)$, range is $(-\frac{\pi}{2}, \frac{\pi}{2})$. For $\cot^{-1}x$: domain is all reals, range is $(0,\pi)$. For $\sec^{-1}x$: domain is $(-\infty,-1] \cup [1,\infty)$, range is $[0,\pi] - \{\frac{\pi}{2}\}$. For $\csc^{-1}x$: domain is $(-\infty,-1] \cup [1,\infty)$, range is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$. The ranges are called principal values - they're chosen to make functions single-valued. When solving equations, all solutions must have arguments in appropriate domains and results in principal value ranges. For example, if equation gives $\sin^{-1}(2)$, this is undefined as $2 > 1$ violates domain. In JEE problems, checking domain/range prevents invalid solutions and earns crucial marks.

To solve equations with inverse trig functions: Step 1 - Use addition/subtraction formulas to combine inverse trig terms into single term. Step 2 - Isolate the inverse trig function on one side. Step 3 - Take appropriate trig function of both sides to remove inverse. For example, if $\tan^{-1}(expression) = \frac{\pi}{6}$, take tan: $expression = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Step 4 - Solve resulting algebraic equation (could be linear, quadratic, or higher). Step 5 - Check all solutions: are they in given interval? Do they satisfy original equation? Are domains satisfied? Common errors include: forgetting to check $ab < 1$ condition before using tan addition formula, not verifying solutions are in required interval, arithmetic mistakes in quadratic formula, incorrect exact values (e.g., confusing $\tan(\frac{\pi}{6})$ with $\tan(\frac{\pi}{4})$). For JEE speed, memorize exact values: $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, $\tan(\frac{\pi}{4}) = 1$, $\tan(\frac{\pi}{3}) = \sqrt{3}$. Practice mental estimation to quickly eliminate impossible solutions.

Memorizing exact trig values for standard angles saves enormous time in JEE. For $\theta = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$: $\sin$ values are $0, \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, 1$. $\cos$ values are reverse: $1, \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}, 0$. $\tan$ values: $0, \frac{1}{\sqrt{3}}, 1, \sqrt{3}, \text{undefined}$. Remember pattern: for sin, numerators go $0,1,2,3,4$ under square roots and divide by 2; for cos, reverse order. For angles like $\frac{5\pi}{6}, \frac{3\pi}{4}$, use reference angles and quadrant signs. For negative angles, use: $\sin(-\theta) = -\sin\theta$, $\cos(-\theta) = \cos\theta$, $\tan(-\theta) = -\tan\theta$. Compound angles: $\sin(\frac{\pi}{12}) = \frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos(\frac{\pi}{12}) = \frac{\sqrt{6}+\sqrt{2}}{4}$ from angle subtraction. Double angles: $\sin(\frac{\pi}{3}) = 2\sin(\frac{\pi}{6})\cos(\frac{\pi}{6})$. These relationships help verify answers and spot errors quickly during exams.