The sum of the infinite series -1 ((7/4)) + -1 ((1 9/4)) + -1 ((3 9/4)) + -1 ((6 7/4)) + . is :-

Solution & Explanation

Related Formula

The general transformation formula for a difference of tangents is:

⁻¹x - ⁻¹y = ⁻¹((x-y)/(1+xy))

Core Logic

Let the general term of the series be $Tₙ$ $T_n$. Examining the numerators ( $7, 19, 39, 67,$ $7, 19, 39, 67, \dots$):

The differences between consecutive terms are $12, 20, 28,$ $12, 20, 28, \dots$, which forms an arithmetic progression with a common difference of $8$ $8$.

Thus, the general term for the sequence in the numerator can be found using difference methods:

Numerator = 4n² + 3

Therefore, the $n$ $n$-th term $Tₙ$ $T_n$ is:

Tₙ = ⁻¹((4n² + 3)/(4)) = ⁻¹((4)/(4n² + 3))

Step 1: Rewriting the general term for telescoping sum

Divide the numerator and denominator inside the argument by $4$ $4$:

Tₙ = ⁻¹((1)/(n² + (3)/(4))) = ⁻¹((1)/(1 + (n² - (1)/(4))))

Factorize $n² - (1)/(4)$ $n^2 - \frac{1}{4}$ as a difference of squares:

Tₙ = ⁻¹(((n + (1)/(2)) - (n - (1)/(2)))/(1 + (n + (1)/(2))(n - (1)/(2))))

Using the difference formula for $⁻¹$ $\tan^{-1}$:

Tₙ = ⁻¹(n + (1)/(2)) - ⁻¹(n - (1)/(2))

Step 2: Telescoping summation

Expanding the sum up to $n$ $n$ terms:

Sₙ = Σk=1ⁿ Tk = [ ⁻¹((3)/(2)) - ⁻¹((1)/(2))] + [ ⁻¹((5)/(2)) - ⁻¹((3)/(2))] + + [ ⁻¹(n + (1)/(2)) - ⁻¹(n - (1)/(2))]

All intermediate terms cancel out, leaving:

Sₙ = ⁻¹(n + (1)/(2)) - ⁻¹((1)/(2))

Step 3: Infinite limit evaluation

Taking the limit as $n \to \infty$ $n \to \infty$:

S_\infty = n \to \infty [ ⁻¹(n + (1)/(2)) - ⁻¹((1)/(2))] = (π)/(2) - ⁻¹((1)/(2))

Pattern Recognition

Whenever you see an infinite series involving $⁻¹$ $\cot^{-1}$ or $⁻¹$ $\tan^{-1}$, try to rearrange the denominator into the form $1 + xy$ $1 + xy$ and check if the numerator matches $x - y$ $x - y$ to set up a standard telescoping structure.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Sequences and Series

More Inverse Trigonometric Functions Previous-Year Questions — Page 9

Q15 jee_main_2024_31_jan_evening Trigonometric Equations

The number of solutions, of the equation

ex - 2e- x = 2

A. $2$ $2$
B. $more than 2$ $\text{more than } 2$
C. $1$ $1$
D. $0$ $0$

Solution

Core Logic

Let $ex = t$ $e^{\sin x} = t$, where $t > 0$ $t > 0$ because exponential functions are strictly positive. Substitute into the equation:

t - (2)/(t) = 2

t² - 2t - 2 = 0

Solve for $t$ $t$ using the quadratic formula:

t = 2 \pm \sqrt(4 - 4(1)(-2))2 = 1 \pm \sqrt(3)

Since $t > 0$ $t > 0$, we discard $1 - \sqrt(3)$ $1 - \sqrt{3}$. Thus, $t = 1 + \sqrt(3) \approx 2.732$ $t = 1 + \sqrt{3} \approx 2.732$. Now, equate back:

ex = 1 + \sqrt(3) x = ln(1 + \sqrt(3))

We know $e \approx 2.718$ $e \approx 2.718$. Since $1 + \sqrt(3) > e$ $1 + \sqrt{3} > e$, it follows that $ln(1 + \sqrt(3)) > 1$ $\ln(1 + \sqrt{3}) > 1$. But the range of $x$ $\sin x$ is $[-1, 1]$ $[-1, 1]$. Therefore, $x$ $\sin x$ cannot equal a value strictly greater than 1. No real solution exists.

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 12 Maths: Continuity and Differentiability

Q16 jee_main_2024_31_jan_evening Properties of ITFs

a = ⁻¹( (5))

and

b = ⁻¹( (5))

, then

a² + b²

is equal to

A. $4π² + 25$ $4\pi^2 + 25$
B. $8π² - 40π + 50$ $8\pi^2 - 40\pi + 50$
C. $4π² - 20π + 50$ $4\pi^2 - 20\pi + 50$
D. $25$ $25$

Solution

Related Formula

⁻¹( x) = x - 2π for x in [3π/2, 5π/2]

⁻¹( x) = 2π - x for x in [π, 2π]

Core Logic

Evaluate $a = ⁻¹( 5)$ $a = \sin^{-1}(\sin 5)$: The principal branch of $⁻¹ x$ $\sin^{-1} x$ is $[-π/2, π/2]$ $[-\pi/2, \pi/2]$. $5$ $5$ radians is approximately $5 \times 57.3^° \approx 286.5^°$ (in 4th quadrant). The equivalent angle in the principal domain is $5 - 2π$ $5 - 2\pi$. Thus, $a = 5 - 2π$ $a = 5 - 2\pi$.

Evaluate $b = ⁻¹( 5)$ $b = \cos^{-1}(\cos 5)$: The principal branch of $⁻¹ x$ $\cos^{-1} x$ is $[0, π]$ $[0, \pi]$. $5$ $5$ radians is in $[π, 2π]$ $[\pi, 2\pi]$. The equivalent angle is $2π - 5$ $2\pi - 5$. Thus, $b = 2π - 5$ $b = 2\pi - 5$.

Calculate $a² + b²$ $a^2 + b^2$:

a² + b² = (5 - 2π)² + (2π - 5)²

$= 2(5 - 2π)²$ $= 2(5 - 2\pi)^2$

= 2(25 + 4π² - 20π)

= 8π² - 40π + 50

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions

Q15 jee_main_2024_31_jan_morning Properties of Inverse Trigonometric Functions

For

α, β, γ \neq 0

. If

⁻¹α + ⁻¹β + ⁻¹γ = π

and

(α + β + γ)(α - γ + β) = 3 αβ

then

γ

equal to

A. $\sqrt(3)2$ $$\frac{\sqrt{3}}{2}$$
B. $1\sqrt(2)$ $$\frac{1}{\sqrt{2}}$$
C. $\sqrt(3) - 12\sqrt(2)$ $$\frac{\sqrt{3} - 1}{2\sqrt{2}}$$
D. $\sqrt(3)$ $\sqrt{3}$

Solution

Core Logic

Let $⁻¹α = A, ⁻¹β = B, ⁻¹γ = C$ . Given $A + B + C = π$ $A + B + C = \pi$. Since $A = α, B = β, C = γ$ , $α, β, γ$ $\alpha, \beta, \gamma$ act like the side lengths of a triangle divided by $2R$ $2R$ by Sine rule. However, directly dealing with the relation:

(α + β + γ)(α + β - γ) = 3αβ

Step 1: Simplify Algebraic Relation

(α + β)² - γ² = 3αβ

α² + β² + 2αβ - γ² = 3αβ

α² + β² - γ² = αβ

Step 2: Triangle Identification

Divide by $2αβ$ $2\alpha\beta$:

(α² + β² - γ²)/(2αβ) = (1)/(2)

By Cosine Rule, $C = (1)/(2)$ $\cos C = \frac{1}{2}$. Since $C = ⁻¹γ$ $C = \sin^{-1}\gamma$, we know $C = γ$ $\sin C = \gamma$. $C = \sqrt(1 - γ²) = (1)/(2)$ .

Step 3: Final Solution

1 - γ² = (1)/(4) γ² = (3)/(4)

Since $C$ $C$ is an angle of a triangle (or sum equals $π$ $\pi$ and elements are positive limits), $γ = C > 0$ $\gamma = \sin C > 0$.

γ = \sqrt(3)2

Pattern Recognition

The expression $(α + β + γ)(α + β - γ) = 3αβ$ perfectly mirrors the Cosine Rule standard form giving $C = 1/2$ $\cos C = 1/2$.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions

The sum of the infinite series $⁻¹ ((7)/(4)) + ⁻¹ ((1 9)/(4)) + ⁻¹ ((3 9)/(4)) + ⁻¹ ((6 7)/(4)) + .$ is :-

Solution & Explanation

Related Formula

Core Logic

Step 1: Rewriting the general term for telescoping sum

Step 2: Telescoping summation

Step 3: Infinite limit evaluation

Pattern Recognition

Chapter Mix

More Inverse Trigonometric Functions Previous-Year Questions — Page 9

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Step 3: Final Solution

Pattern Recognition

Chapter Mix

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The sum of the infinite series ⁻¹ ((7)/(4)) + ⁻¹ ((1 9)/(4)) + ⁻¹ ((3 9)/(4)) + ⁻¹ ((6 7)/(4)) + .$\cot^{-1} \left(\frac {7}{4}\right) + \cot^{-1} \left(\frac {1 9}{4}\right) + \cot^{-1} \left(\frac {3 9}{4}\right) + \cot^{-1} \left(\frac {6 7}{4}\right) + \dots.$ is :-

Solution & Explanation

Related Formula

Core Logic

Step 1: Rewriting the general term for telescoping sum

Step 2: Telescoping summation

Step 3: Infinite limit evaluation

Pattern Recognition

Chapter Mix

More Inverse Trigonometric Functions Previous-Year Questions — Page 9

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Step 3: Final Solution

Pattern Recognition

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The sum of the infinite series $⁻¹ ((7)/(4)) + ⁻¹ ((1 9)/(4)) + ⁻¹ ((3 9)/(4)) + ⁻¹ ((6 7)/(4)) + .$ is :-