Inverse Trigonometric Functions – JEE Main MathematicsPractice Questions & PYQs

To determine the validity of the given statements, we must estimate the ranges for the angles $\alpha$ and $\beta$ by comparing their arguments to known trigonometric values. Since $\frac{6}{11}$ sits between $\frac{1}{2}$ and $\frac{1}{\sqrt{2}}$, the principal value of $\sin^{-1} \left( \frac{6}{11} \right)$ is constrained between $\frac{\pi}{6}$ and $\frac{\pi}{4}$. Multiplying this range by 3 places $\alpha$ strictly between $\frac{\pi}{2}$ and $\frac{3\pi}{4}$. Because this interval lies entirely in the second quadrant, the value of $\cos(\alpha)$ is necessarily negative, confirming the truth of Statement II.

For the second angle, we note that $\frac{4}{9}$ is situated between $0$ and $\frac{1}{2}$, which restricts $\cos^{-1} \left( \frac{4}{9} \right)$ to the interval $(\frac{\pi}{3}, \frac{\pi}{2})$. Multiplying this range by 3 yields a range for $\beta$ of $(\pi, \frac{3\pi}{2})$. Adding the inequalities for $\alpha$ and $\beta$ together, we find that the sum $\alpha + \beta$ must be greater than $\frac{3\pi}{2}$ and less than $\frac{9\pi}{4}$. Since this range places the angle in the fourth quadrant, where the cosine function is positive, Statement I is also valid.

The inverse tangent subtraction identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$ is the essential tool for simplifying the given summation. By observing that the term $\frac{2^{p-1}}{1+2^{2p-1}}$ can be rewritten as $\frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}}$, the summand transforms into the difference $\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$. Performing this sum from $p=1$ to $11$ creates a telescoping series where all intermediate terms cancel out, leaving only the boundary values $\tan^{-1}(2^{11}) - \tan^{-1}(2^0)$. Since $2^0 = 1$ and $\tan^{-1}(1) = \frac{\pi}{4}$, the entire sum simplifies to $\tan^{-1}(2^{11}) - \frac{\pi}{4}$. Adding this result to the $\frac{\pi}{4}$ term outside the summation effectively isolates $\alpha = \tan^{-1}(2^{11})$, meaning $\tan \alpha = 2^{11}$, which equals 2048.

The domain of the function $f(x) = \sin^{-1}(u)$ is defined for $u \in [-1, 1]$, which implies $-1 \le \frac{1}{x^{2-}2x-2} \le 1$. This compound inequality splits into two separate conditions: $\frac{1}{x^{2-}2x-2} \le 1$ and $\frac{1}{x^{2-}2x-2} \ge -1$.

For the first condition, $\frac{1}{x^{2-}2x-2} - 1 \le 0$ simplifies to $\frac{1 - x^2 + 2x + 2}{x^{2-}2x-2} \le 0$, which yields $\frac{-(x^2 - 2x - 3)}{x^{2-}2x-2} \le 0$, or $\frac{x^2 - 2x - 3}{x^{2-}2x-2} \ge 0$. Solving this using the wavy curve method, we obtain $x \in (-\infty , -1] \cup (1-\sqrt{3}, 1+\sqrt{3}) \cup [3, \infty )$.

The second condition, $\frac{1}{x^{2-}2x-2} + 1 \ge 0$, simplifies to $\frac{x^{2-}2x-1}{x^{2-}2x-2} \ge 0$. Solving this inequality, we find $x \in (-\infty , 1-\sqrt{3}) \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup (1+\sqrt{3}, \infty )$.

Intersecting these two intervals results in the domain $x \in (-\infty , -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty )$. Comparing this to the given structure $(-\infty , \alpha] \cup [\beta, \gamma] \cup [\delta, \infty )$, we identify $\alpha = -1$, $\beta = 1-\sqrt{2}$, $\gamma = 1+\sqrt{2}$, and $\delta = 3$. The sum $\alpha + \beta + \gamma + \delta$ is $(-1) + (1-\sqrt{2}) + (1+\sqrt{2}) + 3$, which equals 4.

The expression for $k$ is simplified by setting $\alpha = \frac{1}{2} \cos^{-1}(2/3)$ and $\beta = \frac{1}{2} \sin^{-1}(2/3)$, which implies $\cos 2\alpha = 2/3$ and $\sin 2\beta = 2/3$. Using the half-angle identities $\tan \alpha = \sqrt{\frac{1-\cos 2\alpha}{1+\cos 2\alpha}} = \frac{1}{\sqrt{5}}$ and $\tan \beta = \frac{1-\cos 2\beta}{\sin 2\beta} = \frac{1-\sqrt{5}/3}{2/3} = \frac{3-\sqrt{5}}{2}$, the first term evaluates to $\tan(\frac{\pi}{4} + \alpha) = \frac{1+\tan \alpha}{1-\tan \alpha} = \frac{1+1/\sqrt{5}}{1-1/\sqrt{5}} = \frac{3+\sqrt{5}}{2}$. Summing these results gives $k = \frac{3+\sqrt{5}}{2} + \frac{3-\sqrt{5}}{2} = 3$.

Substituting $k=3$ into the equation $\sin^{-1}(3x-1) = \sin^{-1}x - \cos^{-1}x$ and using the identity $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$ yields $\sin^{-1}(3x-1) = 2\sin^{-1}x - \frac{\pi}{2}$. Taking the sine of both sides results in $3x-1 = \sin(2\sin^{-1}x - \frac{\pi}{2}) = -\cos(2\sin^{-1}x) = -(1-2x^2) = 2x^{2-}1$. Solving the quadratic equation $2x^{2-}3x=0$ provides roots $x=0$ and $x=1.5$, but the condition $3x-1 \in [-1, 1]$ restricts valid solutions to $x \in [0, 2/3]$, confirming $x=0$ as the only solution.

The trigonometric identity for the sum of inverse tangents, $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, allows you to rewrite the equation $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$ as $\tan^{-1}\left(\frac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)}\right) = \frac{\pi}{4}$. Taking the tangent of both sides turns this into $\frac{2 - (\alpha + \beta)}{1 - (1 - \alpha - \beta + \alpha\beta)} = 1$. Simplifying the denominator leads to the expression $\frac{2 - (\alpha + \beta)}{\alpha + \beta - \alpha\beta} = 1$, which equates the numerator to the denominator as $2 - (\alpha + \beta) = \alpha + \beta - \alpha\beta$.

Rearranging the terms gives $2 + \alpha\beta = 2(\alpha + \beta)$. Given that $\beta = \frac{1}{3\alpha}$, the product $\alpha\beta$ is equal to $\frac{1}{3}$. Substituting this value into the equation yields $2 + \frac{1}{3} = 2(\alpha + \beta)$, which simplifies to $\frac{7}{3} = 2(\alpha + \beta)$. Dividing by 2 results in $\alpha + \beta = \frac{7}{6}$, and multiplying this sum by 6 produces a final value of 7.

Converting trigonometric expressions into algebraic forms using right-angled triangle properties reveals the underlying equality. For the expression $\sin(\tan^{-1}(x\sqrt{2}))$, considering a triangle where the tangent is $x\sqrt{2}$ yields an opposite side of $x\sqrt{2}$ and an adjacent side of 1, resulting in a sine value of $\frac{x\sqrt{2}}{\sqrt{2x^{2+}1}}$. For the expression $\cot(\sin^{-1}\sqrt{1-x^2})$, setting the angle $\theta$ such that $\sin \theta = \sqrt{1-x^2}$ implies $\cos \theta = x$, which gives a cotangent value of $\frac{x}{\sqrt{1-x^2}}$. Equating these two expressions allows us to simplify by dividing by $x$ and squaring both sides, resulting in the relation $2(1-x^2) = 2x^{2+}1$. Simplifying this further leads to $2 - 2x^2 = 2x^2 + 1$, which solves to $4x^2 = 1$, and given the domain $x \in (0,1)$, we obtain $x = \frac{1}{2}$.

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ and $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$ are the necessary trigonometric identities. Setting $\alpha = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)$ yields $\tan \alpha = \frac{2}{3}$, and setting $\beta = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$ yields $\tan \beta = \frac{1}{3}$. Using the subtraction identity, $\tan(\alpha - \beta) = \frac{2/3 - 1/3}{1 + (2/3)(1/3)} = \frac{1/3}{11/9} = \frac{3}{11}$. Applying the double-angle formula to the expression $\tan(2(\alpha - \beta))$ results in $\frac{2(3/11)}{1 - (3/11)^2} = \frac{6/11}{1 - 9/121} = \frac{6/11}{112/121} = \frac{33}{56}$.

The formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$ allows the given expression to be simplified to $\tan^{-1} \left( \frac{10x}{1-24x^2} \right) = \frac{\pi}{6}$. Taking the tangent of both sides results in $\frac{10x}{1-24x^2} = \frac{1}{\sqrt{3}}$, which rearranges into the quadratic equation $24x^2 + 10\sqrt{3}x - 1 = 0$. Solving for $x$ using the quadratic formula yields $x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}$. Given the condition $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, which is approximately $-0.204 < x < 0.204$, the negative root $x = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} \approx -0.775$ falls outside this interval, while the positive root $x = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} \approx 0.054$ lies within it. Thus, only one value of $x$ satisfies the equation in the specified domain.

The domain of the function $\cos^{-1}(u)$ requires $-1 \le u \le 1$, and the domain of $\sin^{-1}(v)$ requires $-1 \le v \le 1$. For the first term, setting $-1 \le \frac{2x-5}{11-3x} \le 1$ yields two inequalities: $\frac{2x-5}{11-3x} \ge -1$, which simplifies to $\frac{6-x}{11-3x} \ge 0$, and $\frac{2x-5}{11-3x} \le 1$, which simplifies to $\frac{5x-16}{11-3x} \le 0$. Solving these identifies the intervals $x \in (-\infty , \frac{11}{3}) \cup [6, \infty )$ and $x \in (-\infty , \frac{16}{5}] \cup (\frac{11}{3}, \infty )$, respectively, leading to a combined domain of $x \in (-\infty , 3.2] \cup [6, \infty )$ for the $\cos^{-1}$ term. For the second term, $-1 \le 2x^{2-}3x+1 \le 1$ must hold; the left part $2x^{2-}3x+2 \ge 0$ is satisfied for all real $x$ due to its negative discriminant, while the right part $2x^{2-}3x \le 0$ simplifies to $x \in [0, 1.5]$. Taking the intersection of the domains of both terms, we find the overall domain of $f(x)$ to be $[0, 1.5]$, meaning $\alpha=0$ and $\beta=1.5$. Consequently, $\alpha+2\beta = 0 + 2(1.5) = 3$.

The identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ allows rewriting the expression $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ as $f(t) = 2t^2 - \pi t + \frac{\pi^2}{4}$ where $t = \sin^{-1} x$. Given $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$, the variable $t$ resides in the interval $[-\frac{\pi}{3}, \frac{\pi}{4}]$. The quadratic function $f(t) = 2t^2 - \pi t + \frac{\pi^2}{4}$ represents a parabola opening upward with its vertex at $t = \frac{\pi}{4}$, so the maximum value occurs at the endpoint furthest from the vertex, which is $t = -\frac{\pi}{3}$. Evaluating $f(t)$ at $t = -\frac{\pi}{3}$ yields $2(-\frac{\pi}{3})^2 - \pi(-\frac{\pi}{3}) + \frac{\pi^2}{4} = \frac{2\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \frac{8\pi^2 + 12\pi^2 + 9\pi^2}{36} = \frac{29}{36}\pi^2$. With $m=29$ and $n=36$ coprime, $m+n = 65$.

Simplifying inverse cotangent expressions containing the term $\sqrt{1+\tan^2 \theta}$ requires identifying the quadrant of the angle to resolve the absolute value of $\sec \theta$ and then utilizing trigonometric half-angle identities to reduce the argument to a recognizable cotangent form. For the first term, because 2 radians resides in the second quadrant, $\sec 2$ is negative, which allows us to write the argument as $\frac{-\sec 2 - 1}{\tan 2}$. Utilizing the identity $\frac{1+\cos 2}{\sin 2} = \cot 1$, this transforms the expression into $\cot^{-1}(-\cot 1)$. Applying the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$, this simplifies further to $\pi - \cot^{-1}(\cot 1)$, which yields $\pi - 1$.

For the second term, since $\frac{1}{2}$ radian lies in the first quadrant, $\sec \frac{1}{2}$ is positive, so the expression becomes $\cot^{-1}\left(\frac{\sec \frac{1}{2} + 1}{\tan \frac{1}{2}}\right)$. Simplifying the argument using the identity $\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}} = \cot \frac{1}{4}$ leads to $\cot^{-1}(\cot \frac{1}{4})$, which evaluates to $\frac{1}{4}$. Subtracting this second result from the first term leads to the final value of $\pi - 1 - \frac{1}{4}$, resulting in $\pi - \frac{5}{4}$.

The general term of the series can be expressed as $T_n = \cot^{-1}\left(\frac{4n^{2+}3}{4}\right)$, which is equivalent to $\tan^{-1}\left(\frac{4}{4n^{2+}3}\right)$ for $n \geq 1$. By manipulating the argument, we can write $\frac{4}{4n^{2+}3}$ as $\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n+\frac{1}{2})(n-\frac{1}{2})}$, allowing us to use the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. Substituting these values, the general term becomes $T_n = \tan^{-1}\left(n+\frac{1}{2}\right) - \tan^{-1}\left(n-\frac{1}{2}\right)$, which simplifies further to $\tan^{-1}\left(\frac{2n+1}{2}\right) - \tan^{-1}\left(\frac{2n-1}{2}\right)$.

The sum of the infinite series is the limit of the partial sum $S_N = \sum_{n=1}^N \left(\tan^{-1}\left(\frac{2n+1}{2}\right) - \tan^{-1}\left(\frac{2n-1}{2}\right)\right)$. This is a telescoping series, where most terms cancel out, leaving $S_N = \tan^{-1}\left(\frac{2N+1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)$. Evaluating the limit as $N \to \infty$, the term $\tan^{-1}\left(\frac{2N+1}{2}\right)$ approaches $\frac{\pi}{2}$, resulting in the total sum $S = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{2}\right)$.

The expansion of the cosine of a sum relies on the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$. Applying this to the equation $y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right)$, where $A=\frac{\pi}{3}$ and $B=\cos^{-1}\frac{x}{2}$, involves substituting $\cos A = \frac{1}{2}$, $\sin A = \frac{\sqrt{3}}{2}$, $\cos B = \frac{x}{2}$, and $\sin B = \sqrt{1 - (\frac{x}{2})^2} = \frac{\sqrt{4-x^2}}{2}$. This substitution yields $y = \frac{1}{2}(\frac{x}{2}) - \frac{\sqrt{3}}{2}(\frac{\sqrt{4-x^2}}{2})$.

Multiplying the entire equation by $4$ simplifies the expression to $4y = x - \sqrt{3}\sqrt{4-x^2}$. Rearranging to isolate the radical component gives $4y - x = -\sqrt{3}\sqrt{4-x^2}$. Squaring both sides produces the equation $(4y-x)^2 = 3(4-x^2)$, which expands to $16y^2 + x^2 - 8xy = 12 - 3x^2$. Combining all terms on one side results in $x^2 - 2xy + 4y^2 = 3$. Recognizing that the expression $(x-y)^2 + 3y^2$ expands to $x^2 - 2xy + y^2 + 3y^2$, which simplifies to $x^2 - 2xy + 4y^2$, confirms that the total value is 3.

The identity $\sec^{-1} x + \csc^{-1} x = \frac{\pi}{2}$ holds for all $|x| \ge 1$. Substituting $\csc^{-1} x = \frac{\pi}{2} - \sec^{-1} x$ into the given expression and letting $u = \sec^{-1} x$ yields the function $f(u) = 16 (u^2 + (\frac{\pi}{2} - u)^2) = 32u^2 - 16\pi u + 4\pi^2$. The domain of $\sec^{-1} x$ is $u \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$. On the interval $[0, \frac{\pi}{2})$, $f(u)$ is a parabola opening upward with its vertex at $u = \frac{\pi}{4}$, where it attains a minimum value of $32(\frac{\pi^2}{16}) - 16\pi(\frac{\pi}{4}) + 4\pi^2 = 2\pi^2$. At the endpoints of this interval, $f(u)$ approaches $4\pi^2$. On the interval $(\frac{\pi}{2}, \pi]$, the function is strictly increasing, reaching a maximum value of $f(\pi) = 32\pi^2 - 16\pi^2 + 4\pi^2 = 20\pi^2$. The sum of the absolute maximum value $20\pi^2$ and the absolute minimum value $2\pi^2$ equals $22\pi^2$.

The primary principle relies on the identity $\sin^{-1} x + \sin^{-1} y = \sin^{-1} \left( x \sqrt{1-y^2} + y \sqrt{1-x^2} \right)$, which allows us to consolidate the sum of inverse trigonometric functions. By applying this property to the terms $\sin^{-1} \frac{5}{13}$ and $\sin^{-1} \frac{33}{65}$, we evaluate their sum as:
$\sin^{-1} \left( \frac{5}{13} \sqrt{1 - \left(\frac{33}{65}\right)^2} + \frac{33}{65} \sqrt{1 - \left(\frac{5}{13}\right)^2} \right)$
This calculation simplifies the nested expression to $\sin^{-1} \frac{4}{5}$. Consequently, the entire argument of the cosine function reduces to $\cos \left( \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{4}{5} \right)$. Since $\sin^{-1} \frac{4}{5}$ is equivalent to $\cos^{-1} \frac{3}{5}$, the expression becomes $\cos \left( \sin^{-1} \frac{3}{5} + \cos^{-1} \frac{3}{5} \right)$, which simplifies to $\cos \left( \frac{\pi}{2} \right)$, resulting in a final value of 0.

The identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$ is fundamental to simplifying the given expression by substituting $x = \sin \theta$, which implies $\sqrt{1-x^2} = \cos \theta$ and $\theta = \sin^{-1} x$. Given $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$, the angle $\theta$ resides in the interval $(-\frac{\pi}{6}, \frac{\pi}{4})$. Substituting these values transforms the argument into $\sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} = \sin(\theta + \frac{\pi}{6})$. The resulting expression $\sin^{-1}(\sin(\theta + \frac{\pi}{6}))$ simplifies to $\theta + \frac{\pi}{6}$ because the argument $\theta + \frac{\pi}{6}$ falls within the interval $(0, \frac{5\pi}{12})$, which is contained in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The inverse secant function $\sec^{-1}(u)$ is only defined when the input $u$ is greater than or equal to $1$ or less than or equal to $-1$. For the function $f(x)=\sec^{-1}(2\lfloor x \rfloor+1)$, we apply this condition to the argument $2\lfloor x \rfloor+1$, which leads to the two inequalities $2\lfloor x \rfloor+1 \geq 1$ and $2\lfloor x \rfloor+1 \leq -1$. Simplifying the first expression results in $2\lfloor x \rfloor \geq 0$, which reduces to $\lfloor x \rfloor \geq 0$, while the second expression becomes $2\lfloor x \rfloor \leq -2$, which reduces to $\lfloor x \rfloor \leq -1$. Since $\lfloor x \rfloor \geq 0$ is satisfied by all $x \geq 0$ and $\lfloor x \rfloor \leq -1$ is satisfied by all negative real numbers, every real number satisfies the requirements for the argument, resulting in a domain of $(-\infty , \infty )$.

The relationship between inverse trigonometric functions allows us to rewrite $\cos^{-1} x$ as $\frac{\pi}{2} - \sin^{-1} x$. Substituting this identity into the given equation leads to $\frac{\pi}{2} - \sin^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x+1)$, which simplifies by isolating terms to $-\frac{\pi}{2} - 2\sin^{-1} x = \sin^{-1}(2x+1)$. Applying the sine function to both sides and utilizing the double-angle identity $\cos(2\theta) = 1 - 2\sin^2\theta$, we simplify the equation to $-(1 - 2x^2) = 2x + 1$, which reduces to the quadratic $2x^2 - 2x - 2 = 0$. Solving this quadratic yields the roots $x = \frac{1 \pm \sqrt{5}}{2}$, but given the domain constraints of the inverse sine functions, only $x = \frac{1 - \sqrt{5}}{2}$ is a valid solution. Since this is the only value in the set $S$, the sum $\sum_{x\in S}(2x-1)^2$ is simply $(2(\frac{1 - \sqrt{5}}{2}) - 1)^2$, which equals 5.

Applying the identities $\sec^2 \theta = 1 + \tan^2 \theta$ and $\csc^2 \phi = 1 + \cot^2 \phi$ simplifies the expression $\sec^2(\tan^{-1}\alpha) + \csc^2(\cot^{-1}\beta) = 36$ into $(1 + \tan^2(\tan^{-1}\alpha)) + (1 + \cot^2(\cot^{-1}\beta)) = 36$. This yields $1 + \alpha^2 + 1 + \beta^2 = 36$, which simplifies to $\alpha^2 + \beta^2 = 34$.

Using the provided sum $\alpha + \beta = 8$, we expand $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$ and substitute the known values to obtain $8^2 = 34 + 2\alpha\beta$. Simplifying this gives $64 = 34 + 2\alpha\beta$, which leads to $2\alpha\beta = 30$, or $\alpha \beta = 15$. These conditions define the roots of the quadratic equation $x^2 - 8x + 15 = 0$, which are $x = 3$ and $x = 5$. Since the constraint $\alpha \leq \beta$ must be satisfied, we assign $\alpha = 3$ and $\beta = 5$. Consequently, the requested value is $\alpha^2 + \beta = 3^2 + 5 = 14$.