Inverse Trigonometric Functions – JEE Main MathematicsPractice Questions & PYQs

To simplify the argument within the tangent function, first address the term $2\tan^{-1}(3/5)$ using the identity:
$2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$
Substituting $3/5$ for $x$ results in $\tan^{-1}\left(\frac{6/5}{1-9/25}\right)$, which simplifies to $\tan^{-1}(15/8)$. Next, rewrite $\sin^{-1}(5/13)$ by interpreting it as an angle in a right triangle where the opposite side is 5 and the hypotenuse is 13. Since the adjacent side is $\sqrt{13^2 - 5^2}$, which equals 12, this term is equivalent to $\tan^{-1}(5/12)$.

Combining these two inverse tangent expressions using the formula:
$\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$
yields $\tan^{-1}\left(\frac{15/8 + 5/12}{1-(15/8)(5/12)}\right)$. Calculating the values within the parenthesis involves finding a common denominator, where the numerator becomes $\frac{45+10}{24} = \frac{55}{24}$ and the denominator becomes $\frac{96-75}{96} = \frac{21}{96}$. Multiplying these leads to $\frac{55}{24} \cdot \frac{96}{21} = \frac{220}{21}$. Evaluating the outer tangent function for this result leaves the final value of $\frac{220}{21}$.

The evaluation of $\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$ relies on the recursive application of trigonometric half-angle identities. Defining $\theta = \sin^{-1}\left(\frac{\sqrt{63}}{8}\right)$ gives $\sin\theta = \frac{\sqrt{63}}{8}$, and using the Pythagorean identity yields $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{1}{8}$. Applying the cosine half-angle formula $\cos(\frac{\theta}{2}) = \sqrt{\frac{1+\cos\theta}{2}}$ results in $\cos(\frac{\theta}{2}) = \sqrt{\frac{1+\frac{1}{8}}{2}} = \frac{3}{4}$. Finally, the tangent half-angle identity $\tan(\frac{\alpha}{2}) = \sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}$ with $\alpha = \frac{\theta}{2}$ leads to $\tan(\frac{\theta}{4}) = \sqrt{\frac{1-\frac{3}{4}}{1+\frac{3}{4}}} = \frac{1}{\sqrt{7}}$.

The series expansion for the natural logarithm $\ln(1+x)$ is given by $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$, which is valid for $-1 < x \leq 1$. By regrouping the terms of the given expression $(a+b) - \frac{a^{2}+b^{2}}{2} + \frac{a^{3}+b^{3}}{3} - \frac{a^{4}+b^{4}}{4} + \dots$, we can separate the components involving $a$ and $b$ into two distinct infinite series: $\left( a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \dots \right) + \left( b - \frac{b^2}{2} + \frac{b^3}{3} - \frac{b^4}{4} + \dots \right)$. This simplifies the expression to $\ln(1+a) + \ln(1+b)$, which is equivalent to $\ln((1+a)(1+b))$ using logarithmic properties.

To determine the value of the product $(1+a)(1+b)$, we utilize the trigonometric identity for the sum of inverse tangents:
$\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$
Setting this equal to $\frac{\pi}{4}$ leads to $\frac{a+b}{1-ab} = \tan \left( \frac{\pi}{4} \right) = 1$. This rearrangement yields $a+b = 1-ab$, which implies $a+b+ab = 1$. By adding 1 to both sides, we obtain $1+a+b+ab = 2$, which factors into $(1+a)(1+b) = 2$. Substituting this result into the previous logarithmic expression gives $\ln 2$.

Inverse trigonometric functions return values strictly within their principal domains, so when inputs fall outside these ranges, we must apply symmetry and periodicity properties to identify an equivalent angle within the standard interval. For $\cos^{-1}(\cos(-5))$, we utilize the property $\cos(x) = \cos(2\pi-x)$ to see that $\cos(-5)$ is equivalent to $\cos(5)$, which simplifies to $2\pi-5$ because this value lies within $[0, \pi]$. Regarding $\sin^{-1}(\sin(6))$, the periodicity of sine allows us to write $\sin(6) = \sin(6-2\pi)$, which maps directly to $6-2\pi$ since this value resides within $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Similarly, for $\tan^{-1}(\tan(12))$, shifting the input by multiples of the tangent period $\pi$ gives $12-4\pi$, placing the result within the valid interval of $(-\frac{\pi}{2}, \frac{\pi}{2})$. Combining these transformed values yields the expression $(2\pi-5) + (6-2\pi) - (12-4\pi)$, which simplifies to $4\pi-11$.

The validity of this equation hinges on the strict domain constraints of the inverse trigonometric functions. For $\sin^{-1} \sqrt{x^{2+}x+1}$ to be defined, the argument must satisfy $0 \leq x^{2+}x+1 \leq 1$, which simplifies to $x^{2+}x \leq 0$. Simultaneously, the term $\sqrt{x(x+1)}$ within the $\tan^{-1}$ function requires that $x^{2+}x \geq 0$ for the value to be real. These two conditions can only be satisfied simultaneously if $x^{2+}x = 0$, which occurs specifically at $x = 0$ and $x = -1$. Substituting either of these values into the original equation results in $\tan^{-1}(0) + \sin^{-1}(1)$, which evaluates to $\frac{\pi}{2}$ rather than the required $\frac{\pi}{4}$. Since no real numbers satisfy the equality, there are no real roots.

The equality $\frac{\sin^{-1}(x)}{a}=\frac{\cos^{-1} x}{b}=\frac{\tan^{-1} y}{c}$ allows us to utilize the fundamental identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$. By the properties of proportions, the equality $\frac{\sin^{-1}(x)}{a} = \frac{\cos^{-1}(x)}{b}$ implies each fraction is equivalent to $\frac{\sin^{-1}(x) + \cos^{-1}(x)}{a+b}$, which simplifies to $\frac{\pi}{2(a+b)}$. Equating this result with the third term in the sequence yields $\frac{\tan^{-1}(y)}{c} = \frac{\pi}{2(a+b)}$, which can be rearranged by multiplying both sides by $2$ to produce $2 \tan^{-1}(y) = \frac{\pi c}{a+b}$. Given the trigonometric identity $2 \tan^{-1}(y) = \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right)$, taking the cosine of both sides leads directly to $\cos \left( \frac{\pi c}{a+b} \right) = \frac{1-y^2}{1+y^2}$.

Converting the inverse cotangent term into a more manageable inverse tangent form uses the identity $\cot^{-1}(x) = \tan^{-1}(1/x)$, changing $2 \cot^{-1}(5)$ into $2 \tan^{-1}(1/5)$. Applying the double angle formula $\tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$ where $\theta = 1/5$ leads to:

$\tan^{-1}\left(\frac{2(1/5)}{1-(1/5)^2}\right) = \tan^{-1}\left(\frac{2/5}{24/25}\right) = \tan^{-1}\left(\frac{5}{12}\right)$

The cosecant function now acts on the sum of two angles, $x = \tan^{-1}(5/12)$ and $y = \cos^{-1}(4/5)$. Defining these angles provides the trigonometric ratios $\sin x = 5/13$ and $\cos x = 12/13$ for $x$, and $\sin y = 3/5$ for $y$ based on $\cos y = 4/5$. Calculating the result requires the sine addition identity within the cosecant reciprocal:

$\csc(x+y) = \frac{1}{\sin x \cos y + \cos x \sin y} = \frac{1}{(5/13)(4/5) + (12/13)(3/5)}$

Performing the final arithmetic yields $\frac{65}{56}$.

The domain of a complex function is established by identifying the values that satisfy all internal constraints simultaneously, specifically keeping inverse trigonometric arguments within their valid range of $[-1, 1]$ and ensuring denominators remain non-zero and positive. For the numerator $\cos^{-1}\sqrt{x^{2}-x+1}$, the input must be at most $1$ because the square root is non-negative and the inverse cosine requires an input magnitude no greater than $1$. This leads to $\sqrt{x^{2}-x+1} \leq 1$, which simplifies to $x^{2}-x \leq 0$, restricting $x$ to the interval $[0, 1]$.

For the denominator $\sqrt{\sin^{-1}\left(\frac{2x-1}{2}\right)}$, the argument of the inverse sine must fall within the range $(0, 1]$ to satisfy both the inverse sine's domain and the square root's positivity requirement. The condition $0 < \frac{2x-1}{2} \leq 1$ leads to $0 < 2x - 1 \leq 2$, which rearranges to $1 < 2x \leq 3$, giving the interval $(\frac{1}{2}, \frac{3}{2}]$. Intersecting $[0, 1]$ with $(\frac{1}{2}, \frac{3}{2}]$ yields the final domain $(\frac{1}{2}, 1]$. Given the domain $(\alpha, \beta]$, we have $\alpha = \frac{1}{2}$ and $\beta = 1$, which results in $\alpha + \beta = \frac{3}{2}$.

Utilizing the telescoping series property $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$ allows for a simplified evaluation of this summation. The expression $\tan^{-1} \frac{1}{2r^2}$ is transformed by multiplying the numerator and denominator by 2 to obtain $\tan^{-1} \frac{2}{4r^2}$, which is then rearranged into $\tan^{-1} \frac{(2r+1) - (2r-1)}{1 + (2r+1)(2r-1)}$. This rewrite converts each term into the difference $\tan^{-1}(2r+1) - \tan^{-1}(2r-1)$. Summing this from $r=1$ to $50$ causes intermediate values to cancel, leaving only the boundary terms $\tan^{-1}(101) - \tan^{-1}(1)$. Applying the subtraction identity once more yields $\tan^{-1} \frac{101 - 1}{1 + (101)(1)}$, which simplifies to $\tan^{-1} \frac{50}{51}$, confirming that $\tan p = \frac{50}{51}$.

The function $f(x) = \tan^{-1}(\sin x + \cos x)$ depends on the variation of the expression $\sin x + \cos x$ within the interval $[0, \frac{\pi}{2}]$. This sum can be written as $\sqrt{2} \sin(x + \frac{\pi}{4})$, which attains a minimum of 1 at the boundaries of the interval and a maximum of $\sqrt{2}$ at $x = \frac{\pi}{4}$. Because the inverse tangent function is strictly increasing, the range of the function is defined by $m = \tan^{-1}(1)$ and $M = \tan^{-1}(\sqrt{2})$. Using the trigonometric identity $\tan(M-m) = \frac{\tan M - \tan m}{1 + \tan M \tan m}$, one arrives at the expression $\frac{\sqrt{2} - 1}{1 + \sqrt{2}}$. Multiplying both the numerator and the denominator by the conjugate $\sqrt{2} - 1$ rationalizes the value to $3 - 2\sqrt{2}$.

A telescoping series approach effectively resolves this sum, where each term is expressed as the difference of two inverse trigonometric values. Starting with the general term $T_n = \cot^{-1}(2n^2)$, we first apply the identity $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$, which gives $T_n = \tan^{-1}\left(\frac{1}{2n^2}\right)$. To make use of the subtraction formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$, we adjust the argument by multiplying the numerator and denominator by 2 to obtain $\frac{2}{4n^2}$, which can be rewritten as $\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$.

This transformation allows us to write the $n$-th term as $T_n = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$. As we sum these terms from $n=1$ to $n=100$, the series telescopes such that almost all intermediate terms cancel out, leaving only $\tan^{-1}(201) - \tan^{-1}(1)$. Combining these using the inverse tangent subtraction rule yields $\tan^{-1}\left(\frac{201-1}{1+201 \cdot 1}\right)$, which simplifies to $\tan^{-1}\left(\frac{200}{202}\right)$. Converting this back to the cotangent form results in $\cot^{-1}\left(\frac{202}{200}\right)$, confirming that $\alpha = 1.01$.

The identity $\cot^{-1}\left(\frac{1}{u}\right) = \tan^{-1}(u)$ allows the original equation to be expressed as $\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{8}{31}\right)$. Applying the arctangent addition formula $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ yields $\tan^{-1}\left(\frac{2x}{2-x^2}\right) = \tan^{-1}\left(\frac{8}{31}\right)$, which simplifies to the quadratic equation $8x^2 + 62x - 16 = 0$ after equating the arguments. Dividing by two produces $4x^2 + 31x - 8 = 0$, which factors as $(4x-1)(x+8) = 0$ to provide potential roots of $x = \frac{1}{4}$ and $x = -8$. Testing $x = \frac{1}{4}$ in the original expression results in a value greater than $\frac{\pi}{2}$, which is inconsistent with the right-hand side, rendering this root extraneous. Consequently, the only valid solution is $x = -8$, which is equivalent to $\frac{-32}{4}$.

The domain of the function $\csc^{-1}\left(\frac{1+x}{x}\right)$ is determined by the requirement that the absolute value of its argument must be at least $1$, while also ensuring the denominator $x$ is not zero. By setting $\left|\frac{1+x}{x}\right| \ge 1$ and squaring both sides, we obtain $\frac{(1+x)^2}{x^2} \ge 1$, which simplifies to $1+2x+x^2 \ge x^2$ by multiplying by the positive value $x^2$. This inequality reduces to $2x+1 \ge 0$, yielding $x \ge -\frac{1}{2}$, and excluding $x=0$ provides the final interval $\left[-\frac{1}{2}, \infty \right) \setminus \{0\}$.

To determine the number of solutions, we first analyze the range of the input within the greatest integer functions based on the condition $x \in [-1, 1]$. Since this restriction implies $0 \le x^2 \le 1$, the term $x^2 - \frac{2}{3}$ must fall within the interval $[-\frac{2}{3}, \frac{1}{3}]$. Consequently, the expression $\lfloor x^2 - \frac{2}{3} \rfloor$ can only result in the integer values $-1$ or $0$. Recognizing that $\lfloor x^2 + \frac{1}{3} \rfloor$ is equivalent to $\lfloor x^2 - \frac{2}{3} + 1 \rfloor$, we can apply the property $\lfloor a+1 \rfloor = \lfloor a \rfloor + 1$ to rewrite the equation as $\sin^{-1}(\lfloor x^2 - \frac{2}{3} \rfloor + 1) + \cos^{-1}(\lfloor x^2 - \frac{2}{3} \rfloor) = x^2$.

Setting $k = \lfloor x^2 - \frac{2}{3} \rfloor$, the equation simplifies to $\sin^{-1}(k+1) + \cos^{-1}(k) = x^2$. Evaluating the case where $k = 0$ yields $\sin^{-1}(1) + \cos^{-1}(0) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$, which implies $x^2 = \pi$. This value of $x^2$ is not only outside the permitted range $[0, 1]$ but also inconsistent with the assumption $k = 0$. Similarly, examining $k = -1$ leads to $\sin^{-1}(0) + \cos^{-1}(-1) = 0 + \pi = \pi$, which again results in $x^2 = \pi$. This calculation contradicts both the domain $x^2 \in [0, 1]$ and the specific requirement for $k = -1$, meaning no values of $x$ exist that satisfy the equation.

Determining the domain for functions involving inverse trigonometric terms requires that the argument of each term stays within its valid closed interval of $[-1, 1]$. For the cosine term, we require $-1 \leq \frac{x-1}{x+1} \leq 1$. Solving the right side of this inequality, $\frac{x-1}{x+1} \leq 1$ simplifies to $\frac{x-1}{x+1} - 1 \leq 0$, which yields $\frac{-2}{x+1} \leq 0$, implying $x+1 > 0$ or $x > -1$. The left side, $\frac{x-1}{x+1} \geq -1$, rearranges to $\frac{x-1+x+1}{x+1} \geq 0$, which is $\frac{2x}{x+1} \geq 0$; given $x > -1$, this inequality holds for $x \geq 0$. Thus, the first part of the function is defined for $x \in [0, \infty )$.

Next, we apply the condition $-1 \leq \frac{3x^{2+}x-1}{(x-1)^2} \leq 1$ to the sine term, noting that $x \neq 1$. Breaking this into two parts, the condition $\frac{3x^{2+}x-1}{(x-1)^2} \leq 1$ leads to $3x^{2+}x-1 \leq x^{2-}2x+1$, simplifying to $2x^{2+}3x-2 \leq 0$, or $(2x-1)(x+2) \leq 0$, which restricts $x$ to the interval $[-2, \frac{1}{2}]$. Simultaneously, the inequality $\frac{3x^{2+}x-1}{(x-1)^2} \geq -1$ leads to $3x^{2+}x-1 \geq -(x^{2-}2x+1)$, simplifying to $4x^{2-}x \geq 0$, or $x(4x-1) \geq 0$, which means $x \in (-\infty , 0] \cup [\frac{1}{4}, \infty )$. Combining these restrictions for the sine term gives $x \in [-2, 0] \cup [\frac{1}{4}, \frac{1}{2}]$. Finally, intersecting this with the condition derived from the cosine term, $x \in [0, \infty )$, isolates the value $x = 0$ and the interval $[\frac{1}{4}, \frac{1}{2}]$, which forms the full domain.

The equation $\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1}x$ can be solved using the trigonometric identity $\sin^{-1}A + \sin^{-1}B = \sin^{-1}(A\sqrt{1-B^2} + B\sqrt{1-A^2})$. Substituting the given terms into this formula yields $\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1-\frac{9x^2}{25}} = x$. Simplifying the radicals by accounting for the common denominators leads to the expression $3x\sqrt{25-16x^2} + 4x\sqrt{25-9x^2} = 25x$. Immediately, it is evident that $x=0$ is a valid solution, as it satisfies both sides of the equation.

For values where $x \neq 0$, we can divide both sides by $x$ to obtain $3\sqrt{25-16x^2} + 4\sqrt{25-9x^2} = 25$. Letting $u = 3\sqrt{25-16x^2}$ and $v = 4\sqrt{25-9x^2}$, we are looking to solve the system where $u+v=25$. Squaring the expressions for $u$ and $v$ results in $u^2 = 225-144x^2$ and $v^2 = 400-144x^2$, which implies the difference $v^{2-}u^2 = 175$. Utilizing the algebraic identity $v^{2-}u^2 = (v-u)(v+u)$, we substitute $u+v=25$ to find that $v-u = \frac{175}{25} = 7$. Solving the resulting system of linear equations, $u+v=25$ and $v-u=7$, gives $u=9$ and $v=16$. Substituting $u=9$ back into the definition of $u$ gives $9 = 3\sqrt{25-16x^2}$, which simplifies to $3 = \sqrt{25-16x^2}$. Squaring both sides yields $9 = 25-16x^2$, which results in $16x^2 = 16$, or $x = \pm 1$. Combining these with our earlier result, there are exactly 3 real values for $x$ that satisfy the given equation.

The fundamental relationship $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ serves as the key to this problem by allowing the factorization of the given expression into $(\sin^{-1} x + \cos^{-1} x)(\sin^{-1} x - \cos^{-1} x) = a$. Substituting the known sum identity yields $\frac{\pi}{2}(\sin^{-1} x - \cos^{-1} x) = a$, which simplifies to $\sin^{-1} x - \cos^{-1} x = \frac{2a}{\pi}$. Replacing $\sin^{-1} x$ with $\frac{\pi}{2} - \cos^{-1} x$ transforms the equation into $\frac{\pi}{2} - 2\cos^{-1} x = \frac{2a}{\pi}$, which rearranges into $2\cos^{-1} x = \frac{\pi}{2} - \frac{2a}{\pi}$. Applying the cosine function to both sides results in $\cos(2\cos^{-1} x) = \cos\left(\frac{\pi}{2} - \frac{2a}{\pi}\right)$. By utilizing the double-angle identity $\cos(2\theta) = 2\cos^2 \theta - 1$ where $\theta = \cos^{-1} x$, the left side becomes $2x^2 - 1$, and the right side simplifies to $\sin\left(\frac{2a}{\pi}\right)$ through the property of complementary angles.

Every term in the series can be rewritten using the identity $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ by expressing each general term as $\tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)$, which simplifies to $\tan^{-1}(n+1) - \tan^{-1}(n)$. Applying this to the first ten terms, where $n$ ranges from 1 to 10, transforms the expression into a telescoping sum: $(\tan^{-1}2 - \tan^{-1}1) + (\tan^{-1}3 - \tan^{-1}2) + \dots + (\tan^{-1}11 - \tan^{-1}10)$. Since all intermediate terms cancel, the entire sum reduces to $S = \tan^{-1}11 - \tan^{-1}1$. Calculating $\tan(S)$ requires the tangent subtraction identity $\tan(\tan^{-1}11 - \tan^{-1}1) = \frac{11-1}{1 + 11 \times 1}$, which yields $\frac{10}{12}$, or $\frac{5}{6}$.

The domain of the inverse sine function is restricted to arguments within the closed interval $[-1, 1]$. Because the numerator $|x|+5$ and the denominator $x^{2+}1$ are strictly positive for all real $x$, the left-hand boundary condition is automatically satisfied, leaving only the requirement that $\frac{|x|+5}{x^{2}+1} \le 1$.