Inverse Trigonometric Functions

1 Mark Questions

1. If <math>\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1</math>, then find the

value of x. Delhi 2014

Given, <math>\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1</math> <math>\Rightarrow</math> <math>\sin^{-1}\frac{1}{5} + \cos^{-1}x = \sin^{-1}(1)</math> <math>[: \sin \theta = x \Rightarrow \theta = \sin^{-1} x]</math>

<math display="block">\Rightarrow \sin^{-1}\frac{1}{5} + \cos^{-1}x = \sin^{-1}\left(\sin\frac{\pi}{2}\right) : \sin\frac{\pi}{2} = 1</math> <math display="block">\Rightarrow \sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2}</math> (1/2)

But we know that, <math>\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}</math>,

<math>x \in [-1, 1]</math> <math display="block">\therefore \sin^{-1}\frac{1}{5} = \sin^{-1}x \implies x = \frac{1}{5}</math>

(1/2)

2. If <math>\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}</math>; <math>xy < 1</math>, then write the value of <math>x + y + xy</math>. All India 2014

Given, <math>\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}, xy < 1</math>

We know that,

<math>\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left| \frac{x+y}{1-xy} \right|, xy < 1</math>

<math display="block">\therefore \tan^{-1} \left| \frac{x+y}{1-xy} \right| = \frac{\pi}{4} \implies \frac{x+y}{1-xy} = \tan \frac{\pi}{4} \quad (1/2)</math> <math display="block">x + y = 1 - xy \qquad \left[ \because \tan \frac{\pi}{4} = 1 \right]</math>

<math>x + y + xy = 1</math> (1/2)

3. Write the value of <math>\cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)</math>.

Foreign 2014