( ⁻¹(3)/(5) + ⁻¹(5)/(13) + ⁻¹(33)/(65))$\cos \left(\sin^{-1}\frac{3}{5} +\sin^{-1}\frac{5}{13} +\sin^{-1}\frac{33}{65}\right)$ is equal to:

Related Formula

Core Logic

Track angular positions across quadrants accurately. Evaluate positive/negative absolute value signs based on component radian locations before reducing formulas.

Step 1: Simplify First Exponent Operand

For tracking segment angle height θ = 2$\theta = 2$ radians (Quadrant II), cosine terms switch below zero:

Step 2: Simplify Second Exponent Operand

For tracking segment angle height θ = 1/2$\theta = 1/2$ radian (Quadrant I), expressions remain positive:

Step 3: Combine Structural Terms

Apply inverse mapping functions carefully:

Pattern Recognition

Radian values like 2 sit over 90^°$90^\circ$ but beneath 180^°$180^\circ$. Missing quadrant validation tags is a common trap in inverse identity tracking.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions

Related Formula

The general transformation formula for a difference of tangents is:

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:

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

Step 1: Rewriting the general term for telescoping sum

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

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

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

Step 2: Telescoping summation

Expanding the sum up to n$n$ terms:

All intermediate terms cancel out, leaving:

Step 3: Infinite limit evaluation

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

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

Related Formula

Trigonometric Sine Identity:

Core Logic

Let ⁻¹x = θ x = θ$\sin^{-1}x = \theta \implies x = \sin\theta$ and √(1-x²) = θ$\sqrt{1-x^2} = \cos\theta$. Given constraint -(1)/(2) < x < 1√(2) -(π)/(6) < θ < (π)/(4)$-\frac{1}{2} < x < \frac{1}{\sqrt{2}} \implies -\frac{\pi}{6} < \theta < \frac{\pi}{4}$.

Substitute parameter representations into expression:

Step 1: Check Principal Bounds

Evaluate bounds for arguments: since -(π)/(6) < θ < (π)/(4)$-\frac{\pi}{6} < \theta < \frac{\pi}{4}$:

This lies completely within the principal value branch of ⁻¹x$\sin^{-1}x$, which is [-(π)/(2), (π)/(2)]$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, ⁻¹( (θ + (π)/(6))) = θ + (π)/(6)$\sin^{-1}\left(\sin\left(\theta + \frac{\pi}{6}\right)\right) = \theta + \frac{\pi}{6}$.

Step 2: Final Form

Substituting back θ = ⁻¹x$\theta = \sin^{-1}x$:

Pattern Recognition

Always check primary interval bounds when stripping inverse operators. If the arguments exceed bounds, quadrant mapping transformations must be performed.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions

Related Formula

The standard conversion between ⁻¹(x)$\cot^{-1}(x)$ and ⁻¹(x)$\tan^{-1}(x)$ depends on the sign of x$x$:

Core Logic

Simplify the interior terms algebraic representations:

Step 1: Convert to Inverse Tangent terms

Since α > β > γ > 0$\alpha > \beta > \gamma > 0$:

• (1+αβ)/(α-β) > 0 ⇒ ⁻¹((1+αβ)/(α-β)) = ⁻¹((α-β)/(1+αβ))$\frac{1+\alpha\beta}{\alpha-\beta} > 0 \Rightarrow \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) = \tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)$
• (1+βγ)/(β-γ) > 0 ⇒ ⁻¹((1+βγ)/(β-γ)) = ⁻¹((β-γ)/(1+βγ))$\frac{1+\beta\gamma}{\beta-\gamma} > 0 \Rightarrow \cot^{-1}\left(\frac{1+\beta\gamma}{\beta-\gamma}\right) = \tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right)$
• (1+αγ)/(γ-α) < 0$\frac{1+\alpha\gamma}{\gamma-\alpha} < 0$ (since γ - α < 0$\gamma - \alpha < 0$) ⇒ ⁻¹((1+αγ)/(γ-α)) = π + ⁻¹((γ-α)/(1+αγ))$\Rightarrow \cot^{-1}\left(\frac{1+\alpha\gamma}{\gamma-\alpha}\right) = \pi + \tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right)$

Step 2: Telescopic Sum Evaluation

Apply the difference identity for arctan, ⁻¹((x-y)/(1+xy)) = ⁻¹x - ⁻¹y$\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x - \tan^{-1}y$ :

All variables cancel symmetrically leaving:

= π$= \pi$

Pattern Recognition

The sign trap is the most vital component of this question. The ordering α > β > γ > 0$\alpha > \beta > \gamma > 0$ means the last term contains a denominator with a negative difference (γ - α)$(\gamma - \alpha)$, introducing the +π$+\pi$ offset according to the principal range of ⁻¹(x)$\cot^{-1}(x)$.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions