Inverse Trigonometric Functions Class 12 NCERT Solutions Explained

What Are Inverse Trigonometric Functions?

Inverse trigonometric functions are the inverse relations of the basic trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent. They help find the angle when the value of the trigonometric ratio is known.

Each inverse function is denoted as:

• $\sin^{-1} x$ or $\arcsin x$
• $\cos^{-1} x$ or $\arccos x$
• $\tan^{-1} x$ or $\arctan x$
• $\csc^{-1} x$ or $\arccsc x$
• $\sec^{-1} x$ or $\arcsec x$
• $\cot^{-1} x$ or $\arccot x$

These functions are essential in solving trigonometric equations and calculus problems in Class 12 NCERT syllabus.

Domain and Range of Inverse Trigonometric Functions

Understanding the domain and range is crucial for correctly applying inverse trigonometric functions.

Always check these before solving problems to avoid invalid values.

Important Formulas and Identities

Memorizing key formulas helps simplify complex expressions involving inverse trigonometric functions.

Some important identities include:

• $\sin(\sin^{-1} x) = x$, for $x \in [-1,1]$
• $\cos(\cos^{-1} x) = x$, for $x \in [-1,1]$
• $\tan(\tan^{-1} x) = x$, for all real $x$
• $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$
• $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$
• $\sin^{-1} x = -\sin^{-1}(-x)$

Worked Example:

Find the value of $\sin^{-1} \frac{1}{2} + \cos^{-1} \frac{\sqrt{3}}{2}$.

Solution:

We know $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ and $\cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}$. So,

$$ \sin^{-1} \frac{1}{2} + \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}. $$

Solving Equations Using Inverse Trigonometric Functions

Class 12 NCERT solutions often require solving equations involving inverse trigonometric functions. The general approach involves:

• Isolating the inverse trig function
• Using domain and range restrictions
• Applying identities to simplify

Example:

Solve for $x$:

$$ \sin^{-1} x = \frac{\pi}{6} $$

Solution:

Since $\sin^{-1} x = \frac{\pi}{6}$,

$$ x = \sin \frac{\pi}{6} = \frac{1}{2}. $$

Always verify that the solution lies within the domain of the inverse function.

Applications of Inverse Trigonometric Functions in Class 12

Inverse trigonometric functions are applied in various Class 12 topics such as:

• Calculus: Differentiation and integration of inverse trig functions
• Geometry: Finding angles in triangles when sides are known
• Physics: Resolving components of vectors

For example, the derivative of $y = \sin^{-1} x$ is:

$$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1. $$

This formula is frequently used in calculus problems in the NCERT syllabus.

Tips to Master Inverse Trigonometric Functions for Exams

• Focus on understanding the principal values and branches of each inverse function.
• Practice all NCERT textbook examples and exercises thoroughly.
• Use diagrams to visualize angles and function values.
• Memorize domain and range tables for quick reference.
• Solve previous year question papers to identify common problem types.
• Avoid rote learning; aim for conceptual clarity to tackle application-based questions.