Proving Questions Inverse Trigonometry Functions Lecture 4

Proving Questions Inverse Trigonometry Functions Lecture 4

Advertisements

Part - 1 Lecture - 4 Chapter 2 Inverse Trigonometric Functions

मुश्किल इस दुनिया में कुछ भी नहीं,
फिर भी लोग अपने इरादे तोड़ देते हैं |
अगर सच्चे दिल से हो चाहत कुछ पाने की,
तो सितारे भी अपनी जगह छोड़ देते हैं ||

Booklets/Notes/Assignments are typed here on this website and their PDFs will be made available soon.

Questions discussed in this lecture

Prove:
1. \(3\sin ^{-1} x=\sin ^{-1} (3x-4x^{3} ), x\in \left[-\frac{1}{2} ,\frac{1}{2} \right]\)
(NCERT Exercise 2.2 Q1)

2. \(3\cos ^{-1} x=\cos ^{-1} (4x^{3} -3x), x\in \left[\frac{1}{2} , 1\right]\)
(NCERT Exercise 2.2 Q2)

3. \(\tan ^{-1} \frac{2}{11} +\tan ^{-1} \frac{7}{24} =\tan ^{-1} \frac{1}{2} \)
(NCERT Exercise 2.2 Q3)

4. \(2\tan ^{-1} \frac{1}{2} +\tan ^{-1} \frac{1}{7} =\tan ^{-1} \frac{31}{17} \)
(NCERT Exercise 2.2 Q4)

5. \(\tan ^{-1} \frac{3}{4} +\tan ^{-1} \frac{3}{5} -\tan ^{-1} \frac{8}{19} =\frac{\pi }{4} \)

6. \(\cot ^{-1} 7+\cot ^{-1} 8+\cot ^{-1} 18=\cot ^{-1} 3\)

7. \(\tan ^{-1} x+\tan ^{-1} \left(\frac{2x}{1-x^{2} } \right)=\tan ^{-1} \left(\frac{3x-x^{3} }{1-3x^{2} } \right)\)
(NCERT Example 7)

8. \(\sin ^{-1} (2x\sqrt{1-x^{2} } )=2\sin ^{-1} x=2\cos ^{-1} x\)
(NCERT Example 8, 9)

9. \(\tan ^{-1} \left(\frac{\sqrt{1+x^{2} } +\sqrt{1-x^{2} } }{\sqrt{1+x^{2} } -\sqrt{1-x^{2} } } \right)=\frac{{\pi }}{2} -\frac{1}{2} \sin ^{-1} x^{2} \)

10. \(\tan ^{-1} \sqrt{x} =\frac{1}{2} \cos ^{-1} \left(\frac{1-x}{1+x} \right), x\in [01]\)
(NCERT Miscellaneous Exercise Q9)

Advertisements