Question

The simplified expression of sin ($tan^{-1} x$), for any real number $x$ is given by

• A. $\frac{1}{\sqrt{1+x^{2}}}$
• B. $\frac{x}{\sqrt{1+x^{2}}}$
• C. $-\frac{1}{\sqrt{1+x^{2}}}$
• D. $-\frac{x}{\sqrt{1+x^{2}}}$

Answer 1

The Correct Option is B

Let tan\(^{-1}\) x = \(\theta\) \(\Rightarrow x = tan \theta = \frac{sin \theta }{cos \theta }=\frac{sin \theta }{\sqrt{1-sin^{2} \theta }}\) \(\Rightarrow x =\sqrt{1-sin^{2} \theta }=sin \theta\) \(\Rightarrow x^{2}\left(1-sin^{2} \theta \right)=sin^{2} \theta\) \(\Rightarrow x^{2}=sin^{2} \theta \left(1+x^{2}\right)\) \(\Rightarrow sin^{2} \theta =\frac{x^{2}}{1+x^{2}} \Rightarrow sin \theta =\frac{x}{\sqrt{1+x^{2}}}\) \(\Rightarrow \theta =sin^{-1} \frac{x}{\sqrt{1+x^{2}}}\) \(\Rightarrow tan^{-1} x =sin^{-1} \frac{x}{\sqrt{1+x^{2}}}\) Now, sin \(\left(tan^{-1}x\right) = sin\left( sin^{-1} \frac{x}{\sqrt{1+x^{2}}}\right)\) \(\quad\quad\quad\quad\quad\quad=\frac{x}{\sqrt{1+x^{2}}}\)

Trigonometry is the branch of geometry that explains the connections between a right-angled triangle's angles and sides. It contains identities and formulae that are very useful for computations in maths and science. As was already mentioned, trigonometry also includes ratios and functions like sin, cos, and tan. The issue of what an inverse trigonometric function is may also be answered in the same way.Simply put, inverse trigonometric functions are the opposites of the fundamental trigonometric functions sine, cosine, tangent, cotangent, secant, and cosecant. The terms arcus functions, anti trigonometric functions, and cyclometric functions are also used to describe them. To find the angle for any trigonometric ratio, apply these inverse trigonometric functions. Engineering, physics, geometry, and navigation all heavily rely on the inverse trigonometry functions.