Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #1402
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2025 AISSCE(Board Exam) VSA
Simplify $\sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$.
Prev Next

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Substitution

Let $x = \tan \theta$. This substitution is useful because it allows us to simplify the expression inside the $\sin^{-1}$ function using trigonometric identities.

Step 2: Simplify the expression inside $\sin^{-1}$

Substitute $x = \tan \theta$ into the expression: $$ \frac{x}{\sqrt{1+x^2}} = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} $$

Step 3: Use trigonometric identity

Recall the trigonometric identity: $1 + \tan^2 \theta = \sec^2 \theta$. Therefore, $$ \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} = \frac{\tan \theta}{\sqrt{\sec^2 \theta}} = \frac{\tan \theta}{|\sec \theta|} $$ Assuming $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, $\sec \theta$ is positive, so $|\sec \theta| = \sec \theta$. $$ \frac{\tan \theta}{\sec \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}} = \sin \theta $$

Step 4: Substitute back into $\sin^{-1}$

Now, substitute the simplified expression back into the $\sin^{-1}$ function: $$ \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \sin^{-1}(\sin \theta) = \theta $$

Step 5: Express in terms of $x$

Since we initially let $x = \tan \theta$, we have $\theta = \tan^{-1} x$. Therefore, $$ \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \tan^{-1} x $$

Final Answer: $\tan^{-1} x$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to comprehend the trigonometric identities and apply a suitable substitution to simplify the given expression.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of trigonometric identities and inverse trigonometric functions, which are conceptual knowledge.<\/span>
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of inverse trigonometric functions and their simplification using trigonometric identities, a standard topic in the textbook.

More from this Chapter

MCQ_SINGLE
If \(\tan^{-1}(x^{2}-y^{2})=a\), where 'a' is a constant, then \(\frac{dy}{dx}\) is:
MCQ_SINGLE
\([\sec^{-1}(-\sqrt{2})-\tan^{-1}(\frac{1}{\sqrt{3}})]\) is equal to:
VSA
Draw the graph of $f(x)=\sin^{-1}x, x\in[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$. Also, write range of $f(x)$.
VSA
Find value of k if $sin^{-1}[k~tan(2~cos^{-1}\frac{\sqrt{3}}{2})]=\frac{\pi}{3}.$
VSA
Simplify \(\sin^{-1}(\frac{x}{\sqrt{1+x^{2}}}).\)
View All Questions