Inverse Trigonometric Functions – JEE Main MathematicsPractice Questions & PYQs

The trigonometric identity $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$ allows the original equation $\cos^{-1} \left( \frac{2}{3x} \right) + \cos^{-1} \left( \frac{3}{4x} \right) = \frac{\pi}{2}$ to be rewritten as $\cos^{-1} \left( \frac{3}{4x} \right) = \sin^{-1} \left( \frac{2}{3x} \right)$.

Taking the cosine of both sides leads to the equation $\frac{3}{4x} = \cos \left( \sin^{-1} \left( \frac{2}{3x} \right) \right)$. By applying the identity $\cos(\sin^{-1}(u)) = \sqrt{1 - u^2}$, the expression becomes $\frac{3}{4x} = \sqrt{1 - \left( \frac{2}{3x} \right)^2}$, which simplifies to $\frac{3}{4x} = \frac{\sqrt{9x^2 - 4}}{3x}$.

Multiplying by $3x$ and squaring both sides simplifies the relationship to $\frac{9}{4} = \sqrt{9x^2 - 4}$, and squaring once more results in $\frac{81}{16} = 9x^2 - 4$. Solving for $x^2$ gives $9x^2 = \frac{145}{16}$, which ultimately yields $x = \frac{\sqrt{145}}{12}$.

The problem relies on the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$, which allows us to decompose the given complex summation into a telescoping series. First, simplify the internal term $\sum_{p=1}^{n} 2p$ to $n(n+1)$, turning the expression inside the summation into $\cot^{-1}(1 + n(n+1))$. Using the identity $\cot^{-1}(z) = \tan^{-1}\left(\frac{1}{z}\right)$, this becomes $\tan^{-1}\left(\frac{1}{1 + n(n+1)}\right)$, which is identical to $\tan^{-1}(n+1) - \tan^{-1}(n)$.

Summing this expression from $n=1$ to $19$ creates a series where consecutive terms cancel out, leaving only $\tan^{-1}(20) - \tan^{-1}(1)$. To compute the final cotangent, let $A = \tan^{-1}(20)$ and $B = \tan^{-1}(1)$, then evaluate $\cot(A-B)$ using the formula $\frac{\cot A \cot B + 1}{\cot B - \cot A}$. Since $\cot(\tan^{-1}(20)) = \frac{1}{20}$ and $\cot(\tan^{-1}(1)) = 1$, the final calculation yields $\frac{\left(\frac{1}{20}\right)(1) + 1}{1 - \frac{1}{20}}$, simplifying to the result $\frac{21}{19}$.

Treating the given expression as a quadratic inequality in terms of $\cot^{-1} x$, we can factor $(\cot^{-1} x)^2 - 7(\cot^{-1} x) + 10 > 0$ into $(\cot^{-1} x - 5)(\cot^{-1} x - 2) > 0$. This quadratic inequality holds when the variable is either greater than the larger root or smaller than the smaller root, which leads to the conditions $\cot^{-1} x > 5$ or $\cot^{-1} x < 2$. Because the $\cot^{-1} x$ function is strictly decreasing, applying the cotangent operation to both sides reverses the inequality signs, transforming these conditions into $x < \cot 5$ and $x > \cot 2$, respectively. Combining these two disjoint intervals provides the complete solution set as $(-\infty , \cot 5) \cup (\cot 2, \infty )$.

Applying the inverse tangent addition identity transforms the given equation into $\tan^{-1}(\frac{5x}{1-6x^2}) = \frac{\pi}{4}$. Taking the tangent of both sides simplifies this to the algebraic equation $\frac{5x}{1-6x^2} = 1$, which is equivalent to the quadratic $6x^2 + 5x - 1 = 0$. Solving this equation produces the roots $x = -1$ and $x = \frac{1}{6}$. Since the definition of set $A$ is restricted to non-negative values, the negative result must be discarded, leaving only the single value $x = \frac{1}{6}$ as the valid solution.

To evaluate the expression $\alpha - \beta$, convert the given inverse trigonometric functions into sine and cosine ratios using the properties of a right-angled triangle. For $\alpha = \cos^{-1} \left( \frac{3}{5} \right)$, the adjacent side is $3$ and the hypotenuse is $5$, meaning the opposite side is $4$, which gives $\sin \alpha = \frac{4}{5}$ and $\cos \alpha = \frac{3}{5}$. Similarly, for $\beta = \tan^{-1} \left( \frac{1}{3} \right)$, the opposite side is $1$ and the adjacent side is $3$, so the hypotenuse is $\sqrt{1^2 + 3^2} = \sqrt{10}$, leading to $\sin \beta = \frac{1}{\sqrt{10}}$ and $\cos \beta = \frac{3}{\sqrt{10}}$. Applying the sine subtraction formula $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$, we substitute the values to obtain $\left( \frac{4}{5} \right) \left( \frac{3}{\sqrt{10}} \right) - \left( \frac{3}{5} \right) \left( \frac{1}{\sqrt{10}} \right) = \frac{12 - 3}{5 \sqrt{10}} = \frac{9}{5 \sqrt{10}}$. Therefore, the value of $\alpha - \beta$ is $\sin^{-1} \left( \frac{9}{5 \sqrt{10}} \right)$.

The subtraction identity for inverse sine functions is defined as $\sin^{-1}x - \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right)$. Substituting $x = \frac{12}{13}$ and $y = \frac{3}{5}$ into this formula, we calculate the respective radical terms as $\sqrt{1-(\frac{12}{13})^2} = \frac{5}{13}$ and $\sqrt{1-(\frac{3}{5})^2} = \frac{4}{5}$, which allows us to simplify the expression to $\sin^{-1}\left(\frac{12}{13} \cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{5}{13}\right) = \sin^{-1}\left(\frac{48}{65} - \frac{15}{65}\right) = \sin^{-1}\left(\frac{33}{65}\right)$. Given that $\sin^{-1}\left(\frac{33}{65}\right)$ is equivalent to $\cos^{-1}\left(\frac{56}{65}\right)$ and satisfying the fundamental relation $\sin^{-1}A + \cos^{-1}A = \frac{\pi}{2}$, the expression can be rewritten as $\frac{\pi}{2} - \sin^{-1}\left(\frac{56}{65}\right)$.

The principal branches of inverse trigonometric functions constrain values to the intervals $[-\pi/2, \pi/2]$ for the sine inverse and $[0, \pi]$ for the cosine inverse. Because 10 radians falls outside these standard ranges, we must apply periodic identities to map the input to its corresponding principal value, resulting in $x = \sin^{-1}(\sin 10) = 3\pi - 10$ and $y = \cos^{-1}(\cos 10) = 4\pi - 10$. Subtracting these values leads directly to $y - x = (4\pi - 10) - (3\pi - 10)$, which simplifies to $\pi$.

The relationship between inverse trigonometric functions is best understood by converting them into standard trigonometric form using the cosine difference identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Assigning $A = \cos^{-1} x$ and $B = \cos^{-1} \frac{y}{2}$ allows us to identify $\cos A = x$ and $\cos B = \frac{y}{2}$, which leads to $\sin A = \sqrt{1 - x^2}$ and $\sin B = \sqrt{1 - \frac{y^2}{4}}$. Substituting these into the identity for $\cos(A - B) = \cos \alpha$ results in $x\left(\frac{y}{2}\right) + \sqrt{1 - x^2}\sqrt{1 - \frac{y^2}{4}} = \cos \alpha$.

Isolating the radical term gives $\sqrt{1 - \frac{y^2}{4} - x^2 + \frac{x^2 y^2}{4}} = \cos \alpha - \frac{xy}{2}$. Squaring both sides removes the radical, yielding $1 - x^2 - \frac{y^2}{4} + \frac{x^2 y^2}{4} = \cos^2 \alpha - xy \cos \alpha + \frac{x^2 y^2}{4}$. After canceling the term $\frac{x^2 y^2}{4}$ from both sides, multiplying the entire equation by $4$ transforms it into $4 - 4x^2 - y^2 = 4 \cos^2 \alpha - 4xy \cos \alpha$. Rearranging this to match the requested expression $4x^2 - 4xy \cos \alpha + y^2$ on one side, we get $4 - 4 \cos^2 \alpha = 4x^2 - 4xy \cos \alpha + y^2$. Applying the identity $1 - \cos^2 \alpha = \sin^2 \alpha$, the final expression simplifies to $4 \sin^2 \alpha$.

The identity $\sin^{-1} \left( \frac{2t}{1+t^2} \right) = 2 \tan^{-1} t$ for $|t| \le 1$ allows the function $f(x) = \sin^{-1} \left( \frac{2 \times 3^x}{1+(3^x)^2} \right)$ to be expressed as $f(x) = 2 \tan^{-1}(3^x)$, which is valid since $x = -1/2$ satisfies $3^x \le 1$. Differentiating this expression with respect to $x$ yields $f'(x) = 2 \cdot \frac{1}{1+(3^x)^2} \cdot (3^x \log 3) = \frac{2 \cdot 3^x \log 3}{1+9^x}$. Substituting $x = -1/2$ into the derivative, we compute $f'(-1/2) = \frac{2 \cdot 3^{-1/2} \cdot \log 3}{1+9^{-1/2}} = \frac{2 \cdot (1/\sqrt{3}) \cdot \log 3}{1+1/3} = \frac{2 \log 3}{\sqrt{3}} \cdot \frac{3}{4} = \frac{6 \log 3}{4 \sqrt{3}} = \frac{3 \log 3}{2 \sqrt{3}} = \frac{\sqrt{3} \log 3}{2}$. Applying the logarithmic property $\log 3 = 2 \log \sqrt{3}$, the expression simplifies to $\frac{\sqrt{3} \cdot 2 \log \sqrt{3}}{2} = \sqrt{3} \log \sqrt{3}$.

By utilizing the trigonometric substitution $x^2 = \cos 2\theta$, the expression inside the inverse tangent function can be simplified using fundamental double-angle identities. Replacing the terms with their half-angle equivalents, $1 + \cos 2\theta$ becomes $2\cos^2 \theta$ and $1 - \cos 2\theta$ becomes $2\sin^2 \theta$, which transforms the numerator and denominator into $\sqrt{2}\cos\theta + \sqrt{2}\sin\theta$ and $\sqrt{2}\cos\theta - \sqrt{2}\sin\theta$, respectively. Dividing each term by $\sqrt{2}\cos\theta$ reduces the fraction to the form $\frac{1+\tan\theta}{1-\tan\theta}$, which is equivalent to $\tan\left(\frac{\pi}{4} + \theta\right)$. Consequently, the inverse tangent operation simplifies the entire expression to $\frac{\pi}{4} + \theta$. Applying the relation $\theta = \frac{1}{2} \cos x^2$ leads directly to the result $\frac{\pi}{4} + \frac{1}{2} \cos x^2$.

Inverse trigonometric functions can be converted into algebraic expressions by representing the angle as part of a right-angled triangle where the sides correspond to the trigonometric ratios. For the term $\sin[\cot^{-1}(1+x)]$, let $\theta = \cot^{-1}(1+x)$, which implies $\cot \theta = 1+x$. In a right triangle with adjacent side $1+x$ and opposite side $1$, the hypotenuse is $\sqrt{(1+x)^2 + 1}$, meaning $\sin \theta = \frac{1}{\sqrt{(1+x)^2 + 1}}$. Similarly, for $\cos[\tan^{-1}x]$, let $\phi = \tan^{-1}x$, which implies $\tan \phi = x$. In a right triangle with opposite side $x$ and adjacent side $1$, the hypotenuse is $\sqrt{x^{2+}1}$, meaning $\cos \phi = \frac{1}{\sqrt{x^{2+}1}}$. Equating these values yields $\frac{1}{\sqrt{(1+x)^2 + 1}} = \frac{1}{\sqrt{x^{2+}1}}$. Squaring both sides produces $(1+x)^2 + 1 = x^2 + 1$. Expanding the left side leads to $x^2 + 2x + 1 + 1 = x^2 + 1$, which simplifies to $2x = -1$, resulting in $x = -\frac{1}{2}$.

The addition identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right)$ applies directly here by substituting $A = x$ and $B = \frac{2x}{1-x^2}$, resulting in:
$\tan^{-1}\left(\frac{x + \frac{2x}{1-x^2}}{1 - x\left(\frac{2x}{1-x^2}\right)}\right)$
Simplifying the numerator produces $\frac{3x - x^3}{1-x^2}$, and the denominator reduces to $\frac{1-x^{2-}2x^2}{1-x^2}$, which equals $\frac{1-3x^2}{1-x^2}$. Dividing these two fractions cancels the common $1-x^2$ term, confirming $y = \frac{3x - x^3}{1-3x^2}$ for $|x| < \frac{1}{\sqrt{3}}$.

When three numbers form an arithmetic progression, the middle term must be the arithmetic mean of the other two, represented by the equation $2y = x + z$. If their inverse tangents also constitute an arithmetic progression, they must satisfy the condition $2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z$. Applying the tangent function to both sides and using the sum identity for inverse tangents leads to the equation $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.

By substituting the relationship $2y = x + z$ into this equation, the expressions effectively reduce to the equality $1-y^2 = 1-xz$, which simplifies to $y^2 = xz$. This result indicates that the numbers are simultaneously in an arithmetic progression and a geometric progression. Since the arithmetic mean can only equal the geometric mean when all terms are identical, it follows that $x = y = z$.

Converting the inverse trigonometric components into $\tan^{-1}$ functions simplifies the calculation by enabling the use of standard addition identities. Since $\csc^{-1}\left(\frac{5}{3}\right)$ relates to a right triangle with an opposite side of 3 and a hypotenuse of 5, the adjacent side must be 4, which identifies the term as $\tan^{-1}\left(\frac{3}{4}\right)$. This modifies the original expression to $\cot\left(\tan^{-1}\left(\frac{3}{4}\right)+\tan^{-1}\left(\frac{2}{3}\right)\right)$. Using the addition formula $\tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, the input to the cotangent function becomes $\tan^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}\right)$. Simplifying the numerator and denominator yields $\tan^{-1}\left(\frac{17}{6}\right)$, and taking the cotangent of this inverse tangent value results in $\frac{6}{17}$.

The identity $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ directly relates the sine and cosine inverse functions for complementary angles. Recognizing that $\csc^{-1}(\frac{5}{4})$ is equivalent to $\sin^{-1}(\frac{4}{5})$, we can rewrite the equation as:

$\sin^{-1}\left(\frac{x}{5}\right) + \sin^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2}$

Isolating the sine term on the left side produces $\sin^{-1}(\frac{x}{5}) = \frac{\pi}{2} - \sin^{-1}(\frac{4}{5})$, which is mathematically identical to $\sin^{-1}(\frac{x}{5}) = \cos^{-1}(\frac{4}{5})$. Converting the right side to a sine inverse function involves visualizing a right triangle where the hypotenuse is 5 and the adjacent side is 4, revealing an opposite side of 3. This indicates that $\frac{x}{5} = \sin(\cos^{-1}(\frac{4}{5})) = \frac{3}{5}$, which results in $x = 3$.

The relationship between inverse cosine functions is governed by the identity $\cos^{-1} A - \cos^{-1} B = \cos^{-1}(AB + \sqrt{1-A^2}\sqrt{1-B^2})$. Starting with $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$, taking the cosine of both sides leads to $\frac{xy}{2} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} = \cos \alpha$.

Isolating the square root on one side gives $\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} = \cos \alpha - \frac{xy}{2}$. Squaring both sides produces the equation $(1-x^2)(1-\frac{y^2}{4}) = \cos^2 \alpha - xy \cos \alpha + \frac{x^2 y^2}{4}$. Expanding the left side yields $1 - x^2 - \frac{y^2}{4} + \frac{x^2 y^2}{4}$, which equates to the right side $\cos^2 \alpha - xy \cos \alpha + \frac{x^2 y^2}{4}$. After canceling the $\frac{x^2 y^2}{4}$ term present on both sides, the equation simplifies to $1 - x^2 - \frac{y^2}{4} = \cos^2 \alpha - xy \cos \alpha$. Rearranging this to isolate the trigonometric term results in $x^2 + \frac{y^2}{4} - xy \cos \alpha = 1 - \cos^2 \alpha$, which is equivalent to $x^2 + \frac{y^2}{4} - xy \cos \alpha = \sin^2 \alpha$. Multiplying this entire equation by 4 yields the requested expression $4x^2 + y^2 - 4xy \cos \alpha = 4 \sin^2 \alpha$.

The principal value branch of $\sin^{-1} x$ is restricted to the closed interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since the equation sets $2 \sin^{-1} a$ equal to $\sin^{-1} x$, the expression $2 \sin^{-1} a$ must necessarily fall within this same range, leading to the inequality:
$-\frac{\pi}{2} \leq 2 \sin^{-1} a \leq \frac{\pi}{2}$
Dividing the entire inequality by $2$ results in $-\frac{\pi}{4} \leq \sin^{-1} a \leq \frac{\pi}{4}$. Applying the sine function to each part, we find that $a$ must lie between $\sin(-\frac{\pi}{4})$ and $\sin(\frac{\pi}{4})$, which simplifies to $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$, or more concisely, $|a| \leq \frac{1}{\sqrt{2}}$.

The relationship between inverse trigonometric functions provides a direct path to the solution, specifically the identity $\cot^{-1}(\theta) + \tan^{-1}(\theta) = \frac{\pi}{2}$. By adding this standard property to the given equation $\cot^{-1}(\sqrt{\cos \alpha}) - \tan^{-1}(\sqrt{\cos \alpha}) = x$, the inverse tangent terms cancel out, leaving $2\cot^{-1}(\sqrt{\cos \alpha}) = x + \frac{\pi}{2}$, which simplifies to $\cot^{-1}(\sqrt{\cos \alpha}) = \frac{x}{2} + \frac{\pi}{4}$.

Applying the cotangent function to both sides results in $\sqrt{\cos \alpha} = \cot\left(\frac{x}{2} + \frac{\pi}{4}\right)$, which can be expanded using the sum formula to $\frac{\cot(x/2) - 1}{1 + \cot(x/2)}$. Converting this into sine and cosine functions gives the expression $\frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)}$. Squaring this identity yields $\cos \alpha = \frac{1 - 2\sin(x/2)\cos(x/2)}{1 + 2\sin(x/2)\cos(x/2)}$, which simplifies to $\cos \alpha = \frac{1 - \sin x}{1 + \sin x}$. Rearranging this equation to solve for $\sin x$ gives $\sin x = \frac{1 - \cos \alpha}{1 + \cos \alpha}$, and utilizing the half-angle identity for cosine, $\cos \alpha = \frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)}$, confirms that $\sin x = \tan^2\left(\frac{\alpha}{2}\right)$.

Simplifying the expressions for $\alpha$ and $\beta$ yields the roots required to formulate the quadratic equation. For $\alpha$, consider $\theta = \cos^{-1}(1/\sqrt{5})$, which implies $\cos\theta = 1/\sqrt{5}$ and, by the Pythagorean identity, $\sin\theta = 2/\sqrt{5}$. The argument of the tangent function includes $\sin(2\theta)$, calculated using the double-angle formula $2\sin\theta\cos\theta = 2(2/\sqrt{5})(1/\sqrt{5}) = 4/5$. Substituting this value gives $\alpha = \tan((5\pi/16)(4/5)) = \tan(\pi/4) = 1$.

The value of $\beta = \cos(\sin^{-1}(4/5) + \sec^{-1}(5/3))$ can be determined by noting that $\sin^{-1}(4/5)$ corresponds to an angle with a cosine of $3/5$, and $\sec^{-1}(5/3)$ is equivalent to $\cos^{-1}(3/5)$. Consequently, $\beta = \cos(2\cos^{-1}(3/5))$, which transforms via the identity $\cos(2x) = 2\cos^2 x - 1$ into $2(3/5)^2 - 1 = 18/25 - 1 = -7/25$. With roots $\alpha = 1$ and $\beta = -7/25$, the quadratic equation $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ becomes $x^2 - (1 - 7/25)x - 7/25 = 0$. Multiplying the entire expression by $25$ results in $25x^2 - 18x - 7 = 0$.