Problem 32 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The value of \(\sin ^{-1}\left[\cot \left(\sin ^{-1} \sqrt{\left(\frac{2-\sqrt{3}}{4}\right)}+\cos ^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec ^{-1} \sqrt{2}\right)\right]\) is (a) 0 (b) \(\pi / 4\) (c) \(\pi / 6\) (d) \(\pi / 2\)

Step-by-Step Solution

Answer

None of the options given is correct. The final value doesn't exist within the domain of sine inverse function.

Step 1 - Simplify the given values

First, simplify each of the three inverse trigonometric functions inside the cotangent function: \n\[\sin ^{-1}\left(\sqrt{\left(\frac{2-\sqrt{3}}{4}\right)}\right)+\cos^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec ^{-1} \sqrt{2}\]. \n The value of \(\sin^{-1}\left(\sqrt{\left(\frac{2-\sqrt{3}}{4}\right)}\right)\) can be simplified as \(\frac{\pi}{6}\), the value of \(\cos{-1}\left(\frac{\sqrt{12}}{4}\right)\) is \(\frac{\pi}{2}\), and the \(\sec{-1}\sqrt{2}\) as \(\frac{\pi}{4}\). Hence the three inverse trigonometric functions inside the cotangent function now becomes \(\frac{\pi}{6} + \frac{\pi}{2} + \frac{\pi}{4}\). Calculate the sum to get value \(\frac{11\pi}{12}\).

Step 2 - Apply the cotangent function and inverse sine function

Next, apply the cotangent function to the sum of the resulting values from step 1: \[\cot\left(\frac{11\pi}{12}\right)\]. The cotangent of \(\frac{11\pi}{12}\) is \(-\sqrt{3}\). Then the result within the brackets becomes \(-\sqrt{3}\). The expression now simpler: \[\sin^{-1}\left(-\sqrt{3}\right)\]. Lastly, we calculate this final inverse function to get the end value.

Step 3 - Final calculation and comparison with the options

The final calculation is the sine inverse of \(-\sqrt{3}\), considering the principal values from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), the \(\sin^{-1}(-\sqrt{3})\) doesn't exist. So, we derive that none of the given options (a,b,c,d) is correct since no valid real value can be associated with our final calculation.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in simplifying complex expressions involving trigonometric functions. One common identity is the Pythagorean identity, which relates sine and cosine: \( \sin^2 \theta + \cos^2 \theta = 1 \). This relationship helps in transforming expressions and finding unknown values. When dealing with inverse trigonometric functions like \( \sin^{-1} \) or \( \cos^{-1} \), these identities allow us to navigate between various trigonometric expressions smoothly. By using identities, you can simplify expressions into their most basic forms, making them easier to work with and understand.

Cotangent Function

The cotangent function, denoted as \( \cot \theta \), is the reciprocal of the tangent function. It is defined as \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). Understanding \( \cot \theta \) is crucial because it frequently appears in advanced trigonometric problems, especially when simplifying expressions or solving equations involving multiple trigonometric functions.

• The cotangent function has vertical asymptotes where \( \sin \theta = 0 \), making it undefined at these points.
• Its range is \(( -\infty, \infty )\), and its periodicity is \( \pi \).

In problems involving cotangent, like the one given, you first simplify the arguments using inverse functions, then apply the cotangent function to streamline the expression further.

Principal Values in Trigonometry

Principal values in trigonometry are essential when dealing with inverse trigonometric functions. These values limit the range of the results of inverse functions, ensuring a unique output. For \( \sin^{-1} \), the principal values lie between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). This means that the result of an inverse sine function will always fall within this specific interval.

• For \( \cos^{-1} \), the principal values are from \(0\) to \(\pi\).
• The range for \( \tan^{-1} \) is also \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Knowing these principal values helps in interpreting the results correctly and comparing them against possible answers, as in the given exercise.