JEE Main 2026 (28 January Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$ , then the number of solutions of the equation $sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x$ is ......... .

Q: Why did we reject $x = \frac{3}{2}$? It is a valid algebraic solution!

Algebra is blind to the domain of functions. The function $\sin^{-1}(3x-1)$ is only defined when $-1 \le 3x-1 \le 1$, which means $0 \le x \le \frac{2}{3}$. Since $\frac{3}{2} > \frac{2}{3}$, it simply cannot exist in the world of this function. It is a ghost solution.

Enter Numerical Value:

Visualized Solution

Analyze the Constant $k$

Given: $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$
We need to simplify this expression first.

Substitution for Inverse Trig

Let $α = cos^{- 1} (\frac{2}{3}) ⟹ cos α = \frac{2}{3}$
Using identity: $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$
Therefore, $sin^{- 1} (\frac{2}{3}) = \frac{π}{2} - α$

Rewrite $k$ in terms of $α$

Substitute $α$ into the expression for $k$ :
$k = tan (\frac{π}{4} + \frac{α}{2}) + tan (\frac{1}{2} (\frac{π}{2} - α))$
$k = tan (\frac{π}{4} + \frac{α}{2}) + tan (\frac{π}{4} - \frac{α}{2})$

Apply Tangent Sum Identity

Recall identity: $tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ) = \frac{2}{c o s 2 θ}$
Here, $θ = \frac{α}{2}$ , so $2 θ = α$
$k = \frac{2}{c o s α}$

Calculate Final Value of $k$

We know $cos α = \frac{2}{3}$
$k = \frac{2}{\frac{2}{3}} = 3$
The main equation becomes: $sin^{- 1} (3 x - 1) = sin^{- 1} x - cos^{- 1} x$

Simplify the Right Hand Side

RHS of equation: $sin^{- 1} x - cos^{- 1} x$
Use identity: $cos^{- 1} x = \frac{π}{2} - sin^{- 1} x$
RHS $= sin^{- 1} x - (\frac{π}{2} - sin^{- 1} x) = 2 sin^{- 1} x - \frac{π}{2}$
Equation: $sin^{- 1} (3 x - 1) = 2 sin^{- 1} x - \frac{π}{2}$

Remove Inverse Sine

Take sine on both sides:
$sin (sin^{- 1} (3 x - 1)) = sin (2 sin^{- 1} x - \frac{π}{2})$
LHS simplifies to $3 x - 1$
For RHS, use $sin (θ - \frac{π}{2}) = - cos θ$
$3 x - 1 = - cos (2 sin^{- 1} x)$

Convert to Algebraic Form

Let $θ = sin^{- 1} x ⟹ sin θ = x$
RHS is $- cos (2 θ)$
Use double angle formula: $cos 2 θ = 1 - 2 sin^{2} θ$
$- cos (2 θ) = - (1 - 2 x^{2}) = 2 x^{2} - 1$
Equation becomes: $3 x - 1 = 2 x^{2} - 1$

Visualize the Equations

We need to solve: $3 x - 1 = 2 x^{2} - 1$
LHS represents a straight line: $y = 3 x - 1$
RHS represents a parabola: $y = 2 x^{2} - 1$
The solutions are the intersection points of these two graphs.

Solve the Quadratic Equation

$3 x - 1 = 2 x^{2} - 1$
Cancel $- 1$ from both sides: $3 x = 2 x^{2}$
$2 x^{2} - 3 x = 0$
$x (2 x - 3) = 0$
Possible solutions: $x = 0$ or $x = \frac{3}{2}$

Check Domain Constraints

Original equation has $sin^{- 1} x$ and $sin^{- 1} (3 x - 1)$
Domain of $sin^{- 1} x$ : $x \in [- 1, 1]$
Domain of $sin^{- 1} (3 x - 1)$ : $- 1 \leq 3 x - 1 \leq 1$
$0 \leq 3 x \leq 2 ⟹ x \in [0, \frac{2}{3}]$
Overall valid domain: $x \in [0, \frac{2}{3}]$

Final Conclusion

Valid domain is $x \in [0, \frac{2}{3}]$
Solution $x = 0$ is inside the domain.
Solution $x = \frac{3}{2}$ is outside the domain (rejected).
Therefore, there is only 1 valid solution.

00:00 / 00:00

The Sigma Insight: Solving Inverse Trigonometric Equations

Solution Diagram

Analyzing the Constant $k$

We are given the expression for

k

k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))

Let

α = cos^{- 1} (\frac{2}{3})

, which implies

cos α = \frac{2}{3}

. Using the identity

sin^{- 1} x + cos^{- 1} x = \frac{π}{2}

, we can express the second term as

\frac{1}{2} (\frac{π}{2} - α) = \frac{π}{4} - \frac{α}{2}

The expression for

k

simplifies to:

k = tan (\frac{π}{4} + \frac{α}{2}) + tan (\frac{π}{4} - \frac{α}{2})

Simplifying the Expression

Using the identity

tan (A + B) + tan (A - B) = \frac{2 s i n ( 2 A )}{c o s ( 2 A ) + c o s ( 2 B )}

, or more simply, converting to sine and cosine, we find:

k = \frac{2}{cos α}

Substituting

cos α = \frac{2}{3}

into the equation, we obtain:

k = \frac{2}{2/3} = 3

Solving the Main Equation

Now we address the equation

sin^{- 1} (3 x - 1) = sin^{- 1} x - cos^{- 1} x

. Using the identity

cos^{- 1} x = \frac{π}{2} - sin^{- 1} x

, the right-hand side becomes:

sin^{- 1} (3 x - 1) = 2 sin^{- 1} x - \frac{π}{2}

Taking the sine of both sides:

3 x - 1 = sin (2 sin^{- 1} x - \frac{π}{2})

Applying the property

sin (θ - \frac{π}{2}) = - cos θ

, we get:

3 x - 1 = - cos (2 sin^{- 1} x)

Final Calculation and Domain Check

Using the double angle formula

cos (2 θ) = 1 - 2 sin^{2} θ

, we substitute

θ = sin^{- 1} x

3 x - 1 = - (1 - 2 x^{2})

3 x - 1 = 2 x^{2} - 1

Rearranging into a quadratic equation:

2 x^{2} - 3 x = 0 \Rightarrow x (2 x - 3) = 0

This yields potential solutions

x = 0

and

x = \frac{3}{2}

We must verify these against the domain of

sin^{- 1} (3 x - 1)

, which requires

3 x - 1 \in [- 1, 1]

, or

x \in [0, \frac{2}{3}]

. Since

x = \frac{3}{2}

lies outside this interval, it is an extraneous solution.

The only valid solution is $x = 0$ .

Common Traps & FAQs

Why can't we just square both sides of the equation to get rid of the inverse sine?

Squaring is a dangerous game in trigonometry! It often introduces extraneous roots that do not satisfy the original equation. In inverse trigonometry, the domain is sacred. Always check your final values against the original constraints rather than relying on algebraic manipulation alone.

Why did we reject

x = \frac{3}{2}

? It is a valid algebraic solution!

Algebra is blind to the domain of functions. The function

sin^{- 1} (3 x - 1)

is only defined when

- 1 \leq 3 x - 1 \leq 1

, which means

0 \leq x \leq \frac{2}{3}

. Since

\frac{3}{2} > \frac{2}{3}

, it simply cannot exist in the world of this function. It is a ghost solution.

How did you know to use the

tan (\frac{π}{4} + θ)

identity?

Pattern recognition is key in JEE Advanced. Whenever you see

tan (\frac{π}{4} + \dots)

paired with another tangent term, it is almost always a hint to look for the sum identity. It is a classic way to simplify complex trigonometric expressions.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$ , then the number of solutions of the equation $sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x$ is ......... .

Enter Numerical Value:

Visualized Solution

Analyze the Constant $k$

Substitution for Inverse Trig

Rewrite $k$ in terms of $α$

Apply Tangent Sum Identity

Calculate Final Value of $k$

Simplify the Right Hand Side

Remove Inverse Sine

Convert to Algebraic Form

Visualize the Equations

Solve the Quadratic Equation

Check Domain Constraints

Final Conclusion

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Constant $k$

Simplifying the Expression

Solving the Main Equation

Final Calculation and Domain Check

Similar Questions

The number of solutions of $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{6}$ , where $- \frac{1}{2 6} < x < \frac{1}{2 6}$ , is equal to

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :

The number of real roots of the equation $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $sin^{- 1} (\frac{3 x}{5}) + sin^{- 1} (\frac{4 x}{5}) = sin^{- 1} x$ is equal to:

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $cos^{- 1} (x) - 2 sin^{- 1} (x) = cos^{- 1} (2 x)$ is equal to :

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

Enter Numerical Value:

Visualized Solution

Analyze the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Substitution for Inverse Trig

Rewrite .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Apply Tangent Sum Identity

Calculate Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Simplify the Right Hand Side

Remove Inverse Sine

Convert to Algebraic Form

Visualize the Equations

Solve the Quadratic Equation

Check Domain Constraints

Final Conclusion

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Simplifying the Expression

Solving the Main Equation

Final Calculation and Domain Check

Similar Questions

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Analyze the Constant $k$

Rewrite $k$ in terms of $α$

Calculate Final Value of $k$

Analyzing the Constant $k$

The number of solutions of $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{6}$ , where $- \frac{1}{2 6} < x < \frac{1}{2 6}$ , is equal to

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :

The number of real roots of the equation $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $cos^{- 1} (x) - 2 sin^{- 1} (x) = cos^{- 1} (2 x)$ is equal to :

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is