Inverse Trigonometric Functions – JEE Main MathematicsPractice Questions & PYQs

The simplification relies on the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)$ alongside the property that $\cot^{-1} u = \tan^{-1} (1/u)$ for $u > 0$. First, each algebraic term inside the $\cot^{-1}$ functions simplifies such that $\beta + \frac{1+\beta^2}{\alpha-\beta} = \frac{1+\alpha\beta}{\alpha-\beta}$, $\gamma + \frac{1+\gamma^2}{\beta-\gamma} = \frac{1+\beta\gamma}{\beta-\gamma}$, and $\alpha + \frac{1+\alpha^2}{\gamma-\alpha} = \frac{1+\gamma\alpha}{\gamma-\alpha}$.

Converting these to arctangent terms yields $\tan^{-1} \left(\frac{\alpha-\beta}{1+\alpha\beta}\right) + \tan^{-1} \left(\frac{\beta-\gamma}{1+\beta\gamma}\right) + \pi + \tan^{-1} \left(\frac{\gamma-\alpha}{1+\gamma\alpha}\right)$, where the constant $\pi$ is included for the third term because its argument is negative. Applying the difference identity to each term results in $(\tan^{-1} \alpha - \tan^{-1} \beta) + (\tan^{-1} \beta - \tan^{-1} \gamma) + \pi + (\tan^{-1} \gamma - \tan^{-1} \alpha)$. Canceling the inverse tangent terms leaves a final result of $\pi$.

Simplifying the expression $x^2 + y^2 + 2xy \sin \alpha$ effectively begins by translating the given inverse trigonometric relationship into a manageable algebraic form. Using the substitution $\sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y$, the original equation becomes $\cos^{-1} x + \cos^{-1} y = \frac{\pi}{2} + \alpha$. Taking the cosine of both sides leads to the identity $xy - \sqrt{1-x^2}\sqrt{1-y^2} = \cos\left(\frac{\pi}{2} + \alpha\right)$, which simplifies cleanly to $xy - \sqrt{1-x^2}\sqrt{1-y^2} = -\sin \alpha$.

Rearranging this expression to isolate the radical component yields $xy + \sin \alpha = \sqrt{1-x^2}\sqrt{1-y^2}$. Squaring both sides results in $(xy + \sin \alpha)^2 = (1-x^2)(1-y^2)$, which expands to $x^2 y^2 + 2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2 + x^2 y^2$. Subtracting $x^2 y^2$ from both sides simplifies the equation to $2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2$. Moving the variables to one side reveals that $x^2 + y^2 + 2xy \sin \alpha = 1 - \sin^2 \alpha$, which is equivalent to $\cos^2 \alpha$. Because $\cos^2 \alpha$ must be non-negative, the smallest value it achieves is 0.

The fundamental relationship between inverse trigonometric functions states that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for any value $x$ in the domain $[-1, 1]$. By rewriting the original equation $2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}$ as $2(\sin^{-1} x + \cos^{-1} x) + \cos^{-1} x = \frac{2\pi}{5}$, we can substitute $\frac{\pi}{2}$ for the parenthetical sum to simplify the expression to $\pi + \cos^{-1} x = \frac{2\pi}{5}$. Subtracting $\pi$ from both sides yields $\cos^{-1} x = -\frac{3\pi}{5}$, which falls outside the required principal value range of $[0, \pi]$, indicating that there are zero real solutions.

To solve for $x$ given the equation $\tan^{-1}(x)+\tan^{-1}(2x)=\frac{\pi}{4}$, we first rewrite the expression as $\tan^{-1}(2x) = \frac{\pi}{4} - \tan^{-1}(x)$ and take the tangent of both sides to eliminate the inverse trigonometric functions. Using the identity for the tangent of a difference, the right side simplifies to $\frac{\tan(\frac{\pi}{4}) - \tan(\tan^{-1}(x))}{1 + \tan(\frac{\pi}{4})\tan(\tan^{-1}(x))} = \frac{1-x}{1+x}$, which leads to the equation $2x = \frac{1-x}{1+x}$. Cross-multiplying results in the quadratic equation $2x^{2+}3x-1=0$. Since we are specifically searching for positive real values, we analyze the roots of this quadratic and find that it yields only one positive value for $x$. Therefore, there is 1 positive real value satisfying the equation.

The double angle identity $\cos(2\theta) = 1 - 2\sin^2 \theta$ relates the trigonometric expression to the variable $x$ by substituting $\theta = \sin^{-1} x$. Setting $\cos(2\theta) = \frac{1}{9}$ leads to the equation $1 - 2x^2 = \frac{1}{9}$, which simplifies to $2x^2 = \frac{8}{9}$ or $x^2 = \frac{4}{9}$. Because $m$ and $n$ are natural numbers, $x = \frac{2}{3}$ identifies $m = 2$ and $n = 3$.

Substituting these values into the expression $mx^2 - nx - m + n = 0$ produces $2x^2 - 3x - 2 + 3 = 0$, which simplifies to $2x^2 - 3x + 1 = 0$. Factoring this quadratic equation as $(2x - 1)(x - 1) = 0$ yields the roots $x = 1$ and $x = \frac{1}{2}$. Given the condition $\alpha > \beta$, we assign $\alpha = 1$ and $\beta = \frac{1}{2}$. Checking the point $(1, \frac{1}{2})$ against the options confirms it satisfies the linear equation $5x + 8y = 9$, as $5(1) + 8(\frac{1}{2}) = 9$.

The property $\cot^{-1} x = \tan^{-1} \frac{1}{x}$ is essential for converting the given expression into a sum of tangent terms, yielding:
$\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{n} = \frac{\pi}{4}$
Combining the first three terms using the addition formula for inverse tangents simplifies the expression to $\tan^{-1} \frac{23}{24} + \tan^{-1} \frac{1}{n} = \frac{\pi}{4}$. Isolating the term with $n$ results in $\tan^{-1} \frac{1}{n} = \tan^{-1} 1 - \tan^{-1} \frac{23}{24}$. Applying the difference identity $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \frac{a-b}{1+ab}$ gives $\tan^{-1} \frac{1}{n} = \tan^{-1} \frac{1 - \frac{23}{24}}{1 + \frac{23}{24}}$, which simplifies to $\tan^{-1} \frac{1}{47}$. Consequently, $n = 47$.

Inverse trigonometric functions operate within restricted principal value ranges, requiring us to map inputs back into these intervals to identify their equivalent values. For $a = \sin^{-1}(\sin 5)$, since 5 lies in the fourth quadrant, we utilize the relation $\sin(5) = \sin(5 - 2\pi)$ to find $a = 5 - 2\pi$, while for $b = \cos^{-1}(\cos 5)$, we apply $\cos(5) = \cos(2\pi - 5)$ to determine $b = 2\pi - 5$ because 5 exceeds the range $[0, \pi]$. Squaring both values gives $a^2 = (5 - 2\pi)^2$ and $b^2 = (2\pi - 5)^2$, which both expand to $4\pi^2 - 20\pi + 25$. Summing these expressions leads to $a^2 + b^2 = 8\pi^2 - 40\pi + 50$.

Defining $A = \sin^{-1} \alpha$, $B = \sin^{-1} \beta$, and $C = \sin^{-1} \gamma$ implies $\alpha = \sin A$, $\beta = \sin B$, and $\gamma = \sin C$ where $A + B + C = \pi$. Expanding the relation $(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3 \alpha \beta$ results in $(\alpha + \beta)^2 - \gamma^2 = 3 \alpha \beta$, which simplifies to $\alpha^2 + \beta^2 - \gamma^2 = \alpha \beta$. Dividing both sides by $2 \alpha \beta$ yields $\frac{\alpha^2 + \beta^2 - \gamma^2}{2 \alpha \beta} = \frac{1}{2}$, which equates to $\cos C = \frac{1}{2}$ based on the cosine rule. Applying the identity $\gamma = \sin C = \sqrt{1 - \cos^2 C}$ gives $\gamma = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$.

The domain of this function depends on satisfying the specific constraints for both the inverse sine and logarithmic components. For the inverse sine portion, the internal ratio must lie within the interval $[-1, 1]$, requiring $-1 \le \frac{3x-22}{2x-19} \le 1$, while the logarithmic portion necessitates that the argument remains strictly positive, which means $\frac{3x^{2-}8x+5}{x^{2-}3x-10} > 0$.

Intersecting the solutions to these two inequalities provides the valid domain $\left(5, \frac{41}{5}\right]$. Comparing this to the given interval $(\alpha, \beta]$ reveals that $\alpha = 5$ and $\beta = \frac{41}{5}$. Substituting these values into $3\alpha + 10\beta$ gives $3(5) + 10\left(\frac{41}{5}\right)$, which simplifies to $15 + 82$, resulting in a final value of 97.

Trigonometric substitution effectively simplifies inverse functions when arguments take recognizable forms. Substituting $x = \tan \theta$ with $\theta$ constrained to the interval $(0, \pi/4)$ allows the left side of the equation to be expressed as $2 \tan^{-1} (\tan (\frac{\pi}{4} - \theta))$, which simplifies to $2(\frac{\pi}{4} - \theta)$, or $\frac{\pi}{2} - 2\theta$. Simultaneously, the right side becomes $\cos^{-1} (\cos 2\theta)$, which resolves to $2\theta$ because the specified range of $\theta$ keeps $2\theta$ within the principal domain of the inverse cosine function. Equating these results gives $\frac{\pi}{2} - 2\theta = 2\theta$, which simplifies to $4\theta = \frac{\pi}{2}$, or $\theta = \frac{\pi}{8}$. The only valid solution is therefore $x = \tan \frac{\pi}{8} = \sqrt{2} - 1$, a value approximately equal to 0.414, confirming that there is exactly one element in $S$ and it is strictly less than $\frac{1}{2}$.

The identity $2 \tan^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$ allows the equation $\sin^{-1} x = 2 \tan^{-1} x$ to be rewritten as $x = \frac{2x}{1+x^2}$. This expression simplifies to $x \left( \frac{x^{2-}1}{1+x^2} \right) = 0$, which provides the possible values $x = 0$, $x = 1$, and $x = -1$. Given the constraint $x \in (-1, 1]$, the value $x = -1$ falls outside the specified interval and must be rejected, resulting in 2 solutions.

The inequality $\sin^{-1}(\sin\theta) - \cos^{-1}(\sin\theta) > 0$ is resolved using the identity $\cos^{-1}(\sin\theta) = \frac{\pi}{2} - \sin^{-1}(\sin\theta)$, which simplifies the expression to $2\sin^{-1}(\sin\theta) > \frac{\pi}{2}$, or $\sin^{-1}(\sin\theta) > \frac{\pi}{4}$. For $\theta \in (0, 2\pi)$, this condition is satisfied when $\theta \in (\frac{\pi}{4}, \frac{3\pi}{4})$, which defines the interval $(a, b) = (\frac{\pi}{4}, \frac{3\pi}{4})$ and establishes that $b - a = \frac{\pi}{2}$.

The equation $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$ is defined only when the argument of the inverse trigonometric functions, $x^2 - 6x + 10$, falls within the range $[-1, 1]$. Since $x^2 - 6x + 10 = (x-3)^2 + 1$, the only way this value remains $\leq 1$ is if $(x-3)^2 = 0$, implying $x = 3$. Substituting $x = 3$ and $\beta = \alpha - \frac{\pi}{2}$ into the original quadratic expression leads to $9\alpha + 3(\alpha - \frac{\pi}{2}) + \frac{\pi}{2} = 0$, which simplifies to $12\alpha - \pi = 0$, giving $\alpha = \frac{\pi}{12}$.

The domain of a complex function defined by the sum of multiple terms is determined by finding the intersection of the validity intervals for each component. The logarithmic term $\ln(4x^{2+}11x+6)$ is defined only where the argument is strictly positive, which means $4x^{2+}11x+6 > 0$. Factoring this expression gives $(4x+3)(x+2) > 0$, resulting in the valid intervals $x \in (-\infty , -2) \cup (-3/4, \infty )$. For the inverse sine term $\sin^{-1}(4x+3)$, the input must satisfy $-1 \leq 4x+3 \leq 1$, which simplifies to $4x \in [-4, -2]$ and gives the interval $x \in [-1, -1/2]$. The inverse cosine term $\cos^{-1}\left(\frac{10x+6}{3}\right)$ requires its argument to be within $[-1, 1]$, leading to $-1 \leq \frac{10x+6}{3} \leq 1$, which becomes $-3 \leq 10x+6 \leq 3$, yielding $x \in [-9/10, -3/10]$.

Determining the common region shared by the intervals $(-\infty , -2) \cup (-3/4, \infty )$, $[-1, -1/2]$, and $[-9/10, -3/10]$ requires finding the overlap where all conditions are met simultaneously. Comparing these intervals, the only shared range is $(-3/4, -1/2]$, which identifies the parameters as $\alpha = -3/4$ and $\beta = -1/2$. Calculating the sum $\alpha+\beta$ results in $-3/4 - 1/2 = -5/4$. The final value requested is $36|\alpha+\beta|$, which is $36 \times |-5/4|$, equaling $45$.

The problem hinges on the property $\cot^{-1}\left(\frac{1-x^2}{2x}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ for $x(1-x^2) > 0$ and $\cot^{-1}\left(\frac{1-x^2}{2x}\right) = \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ when $x(1-x^2) < 0$. With $x \in (-1, 1)$ and $x \neq 0$, the positive interval $x \in (0, 1)$ simplifies the given equation to $2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3}$. Setting $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{6}$ leads to the equation $\frac{2x}{1-x^2} = \frac{1}{\sqrt{3}}$, which rearranges to $x^2 + 2\sqrt{3}x - 1 = 0$, yielding the valid root $x = 2 - \sqrt{3}$.

In the negative interval $x \in (-1, 0)$, the identity includes an additional $\pi$ term, transforming the equation into $\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3}$. This reduces to $2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = -\frac{2\pi}{3}$, which means $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = -\frac{\pi}{3}$. Equating $\frac{2x}{1-x^2} = -\sqrt{3}$ results in $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$, where the only solution within the range $x \in (-1, 0)$ is $x = -\frac{1}{\sqrt{3}}$. Summing these two identified values, $2 - \sqrt{3} - \frac{1}{\sqrt{3}}$ becomes $2 - \frac{4}{\sqrt{3}}$, confirming that $\alpha$ is 2.

Rearranging the equation $\sin^{-1} \frac{\alpha}{17} + \cos^{-1} \frac{4}{5} - \tan^{-1} \frac{77}{36} = 0$ yields $\sin^{-1} \frac{\alpha}{17} = \tan^{-1} \frac{77}{36} - \cos^{-1} \frac{4}{5}$. Since $\cos^{-1} \frac{4}{5}$ is equivalent to $\tan^{-1} \frac{3}{4}$, applying the subtraction identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$ results in $\sin^{-1} \frac{\alpha}{17} = \tan^{-1} \frac{8}{15}$. Recognizing that a right triangle with legs $8$ and $15$ has a hypotenuse of $17$, this equates to $\sin^{-1} \frac{8}{17}$, confirming that $\alpha = 8$. Because $8$ radians lies within the interval $(2\pi, 3\pi)$, the expression $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8)$ simplifies using standard periodicity properties to $(3\pi - 8) + (8 - 2\pi)$, which results in a final value of $\pi$.

Recognizing that the terms within the inverse sine functions correspond to angles whose tangents are $x+1$ and $x$, respectively, allows for a direct algebraic simplification. By letting $\theta_1 = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^{2+}1}}\right)$ and $\theta_2 = \sin^{-1}\left(\frac{x}{\sqrt{x^{2+}1}}\right)$, the given equation $\theta_1 - \theta_2 = \frac{\pi}{4}$ implies that $\tan(\theta_1 - \theta_2) = 1$. Utilizing the tangent subtraction identity, $\frac{\tan(\theta_1) - \tan(\theta_2)}{1 + \tan(\theta_1)\tan(\theta_2)} = 1$, we substitute $\tan(\theta_1) = x+1$ and $\tan(\theta_2) = x$ to obtain $\frac{(x+1) - x}{1 + (x+1)x} = 1$. This simplifies to $1 = 1 + x^2 + x$, which reduces to $x^2 + x = 0$, yielding the solutions $x = 0$ and $x = -1$. With the set $S$ identified as $\{0, -1\}$, we proceed to calculate the summation by substituting these values into the expression. For both $x=0$ and $x=-1$, the value of $x^{2+}x+5$ is $5$, leading to the term $\sin\left(\frac{5\pi}{2}\right) - \cos(5\pi)$, which evaluates to $1 - (-1) = 2$. Adding these results for both members of the set $S$ gives a total of $4$.

The domain of the function $\sec^{-1}(u)$ is restricted by the condition $|u| \ge 1$, as this defines the valid input range for the arcsecant operation. Applying this to $\left| \frac{2x}{5x+3} \right| \ge 1$ requires that the denominator $5x+3$ not equal zero, establishing the exclusion $x \neq -3/5$.

Squaring both sides of the inequality yields $(2x)^2 \ge (5x+3)^2$, which rearranges into $(2x)^2 - (5x+3)^2 \ge 0$. Factoring the difference of squares gives $(2x - 5x - 3)(2x + 5x + 3) \ge 0$, simplifying to $(-3x - 3)(7x + 3) \ge 0$. Multiplying by $-1$ flips the inequality, resulting in $3(x+1)(7x+3) \le 0$, which identifies the interval $[-1, -3/7]$. Accounting for the point where the function is undefined, the domain becomes $[-1, -3/5) \cup (-3/5, -3/7]$.

Matching these endpoints to the form $[\alpha, \beta) \cup (\gamma, \delta]$ identifies $\alpha = -1$, $\beta = -3/5$, $\gamma = -3/5$, and $\delta = -3/7$. Calculating the expression $|3\alpha+10(\beta+\gamma)+21\delta|$ requires substituting these values to get $|3(-1) + 10(-3/5 - 3/5) + 21(-3/7)|$, which simplifies to $|-3 + 10(-6/5) - 9|$. This evaluates to $|-3 - 12 - 9|$, which is $|-24|$, resulting in 24.

The core of this problem lies in applying the identity for the difference of two inverse tangents, which is defined as $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. Since the sequence $a_1, a_2, \ldots, a_{2022}$ consists of consecutive natural numbers, the difference between any two adjacent terms is always $1$, meaning $a_{k+1} - a_k = 1$ for every $k$. We can substitute this relationship into the general term of the series, transforming each element $\tan^{-1}\left(\frac{1}{1+ a_k a_{k+1}}\right)$ into $\tan^{-1}\left(\frac{a_{k+1} - a_k}{1 + a_k a_{k+1}}\right)$, which is equivalent to $\tan^{-1}(a_{k+1}) - \tan^{-1}(a_k)$.

Expanding the summation using this identity creates a telescoping series, where every term cancels out with the subsequent one except for the first and last components. This leaves us with the simplified expression $\tan^{-1}(a_{2022}) - \tan^{-1}(a_1)$. Given that the sequence starts with $a_1 = 1$, we know $\tan^{-1}(1) = \frac{\pi}{4}$, and since the numbers are consecutive natural numbers, the $2022^{\text{nd}}$ term is simply $2022$. Substituting these values into our expression gives the final result of $\tan^{-1}(2022) - \frac{\pi}{4}$.

Simplifying the trigonometric arguments reveals that both components of the expression correspond to standard angles. In the first term, factoring $\sqrt{3}$ from the denominator $3+\sqrt{3}$ as $\sqrt{3}(\sqrt{3}+1)$ allows the $1+\sqrt{3}$ term to cancel, reducing the expression to $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$, which equals $\frac{\pi}{6}$. The second term, $\sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$, simplifies by factoring 4 from the numerator and 3 from the denominator within the square root to become $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$, which also equals $\frac{\pi}{6}$. Summing these values yields $\frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$.