Lecture-4

Type – I
\sin \theta =\frac{1}{{\rm cosec}\, \theta }

{\rm cosec\; }\theta =\frac{1}{\sin \theta }

\cos \theta =\frac{1}{\sec \theta }

\sec \theta =\frac{1}{\cos \theta }

\tan \theta =\frac{1}{\cot \theta } =\frac{\sin \theta }{\cos \theta }

\cot \theta =\frac{1}{\tan \theta } =\frac{\cos \theta }{\sin \theta }

Type – II

1=\sin ^{2} \theta +\cos ^{2} \theta

1-\sin ^{2} \theta =\cos ^{2} \theta

1-\cos ^{2} \theta =\sin ^{2} \theta

{\rm cosec}^{2} \theta =1+\cot ^{2} \theta

{\rm cosec}^{2} \theta -\cot ^{2} \theta =1

{\rm cosec}^{2} \theta -1=\cot ^{2} \theta

\sec ^{2} \theta =1+\tan ^{2} \theta

\sec ^{2} \theta -\tan ^{2} \theta =1

\sec ^{2} \theta -1=\tan ^{2} \theta

Type – III

\sin (A+B)=\sin A\cos B+\cos A\sin B

\cos (A+B)=\cos A\cos B-\sin A\sin B

\cos (A-B)=\cos A\cos B+\sin A\sin B

\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}

\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}

\cot (A+B)=\frac{\cot A\cot B-1}{\cot B+\cot A}

\cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}

Type – IV

\sin (A+B)+\sin (A-B)=2\sin A\cos B

\sin (A+B)-\sin (A-B)=2\cos A\sin B

\cos (A+B)+\cos (A-B)=2\cos A\cos B

\cos (A+B)-\cos (A-B)=-2\sin A\sin B

Type – V

\sin 3\, \theta \, =3\sin \theta -4\sin ^{3} \theta

\cos 3\, \theta =4\cos ^{3} \theta -3\cos \theta

\tan 3\, \theta =\frac{3\tan \theta -\tan ^{3} \theta }{1-3\tan ^{2} \theta }

Type – VI

1-\cos \theta =2\sin ^{2} \frac{\theta }{2}

1+\cos \theta =2\cos ^{2} \frac{\theta }{2}

\tan \frac{\theta }{2} \, \, =\, \, \sqrt{\frac{1-\cos \theta }{1+\cos \theta } }

\frac{1-\tan \theta }{1+\tan \theta } =\tan \left(\frac{\pi }{4} -\theta \right)

\frac{1+\tan \theta }{1-\tan \theta } =\tan \left(\frac{\pi }{4} +\theta \right)

1 + sin\theta = \left ( \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \right ) ^2

1 – sin\theta = \left ( \cos \frac{\theta}{2} – \sin \frac{\theta}{2} \right ) ^2

Type – VII

\sin A+\sin B=2\sin \left(\frac{A+B}{2} \right)\, \, \cos \left(\frac{A-B}{2} \right)

\sin A-\sin B=2\cos \left(\frac{A+B}{2} \right)\, \, \sin \left(\frac{A-B}{2} \right)

\cos A+\cos B=2\cos \left(\frac{A+B}{2} \right)\, \, \cos \left(\frac{A-B}{2} \right)

\cos A-\cos B=-2\sin \left(\frac{A+B}{2} \right)\, \, \sin \left(\frac{A-B}{2} \right)

Type – VIII

\sin 2\, \theta =2\sin \theta \cos \theta =\frac{2\tan \theta }{1+\tan ^{2} \theta }

\cos 2\, \theta =\cos ^{2} \theta -\sin ^{2} \theta \, =2\cos ^{2} \theta -1\, =1-2\sin ^{2} \theta \, =\frac{1-\tan ^{2} \theta }{1+\tan ^{2} \theta }

\tan 2\, \theta =\frac{2\tan \theta }{1-\tan ^{2} \theta }

Type – IX

\tan ^{-1} A+\tan ^{-1} B=\tan ^{-1} \left(\frac{A+B}{1-AB} \right)

\tan ^{-1} A-\tan ^{-1} B=\tan ^{-1} \left(\frac{A-B}{1+AB} \right)

Type – X

sin^{-1}\theta + cos^{-1}\theta =  \frac{\pi}{2}

tan^{-1}\theta + cot^{-1}\theta =  \frac{\pi}{2}

sec^{-1}\theta + cosec^{-1}\theta =  \frac{\pi}{2}

Type – XI

sin^{-1}(-\theta)=  -sin^{-1 }\theta

tan^{-1}(-\theta)=  -tan^{-1 }\theta

cosec^{-1}(-\theta)=  -cosec^{-1 }\theta

cos^{-1}(-\theta)= \pi – cos^{-1 }\theta

sec^{-1}(-\theta)= \pi – sec^{-1 }\theta

cot^{-1}(-\theta)= \pi – cot^{-1 }\theta

Type – XII

sin^{-1}\theta = cosec^{-1}{\frac{1}{\theta}}

cosec^{-1}\theta = sin^{-1}{\frac{1}{\theta}}

cos^{-1}\theta = sec^{-1}{\frac{1}{\theta}}

sec^{-1}\theta = cos^{-1}{\frac{1}{\theta}}

tan^{-1}\theta = cot^{-1}{\frac{1}{\theta}}

cot^{-1}\theta = tan^{-1}{\frac{1}{\theta}}

Type – XIII

2\tan ^{-1} \theta =\sin ^{-1} \left(\frac{2\theta }{1+\theta ^{2} } \right)\, \, =\, \, \cos ^{-1} \left(\frac{1-\theta ^{2} }{1+\theta ^{2} } \right)\, \, \, =\, \, \tan ^{-1} \left(\frac{2\theta }{1-\theta ^{2} } \right)

PART - 1

Inverse Trigonometric Functions Lecture 4 Part 1
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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{{\rm \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 [0,\, 1] (NCERT Miscellaneous Exercise Q9)

PART - 2

Inverse Trigonometric Functions Lecture 4 Part 2
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Questions discussed in this lecture

Prove:
11. \frac{9\pi }{8} -\frac{9}{4} \sin ^{-1} \frac{1}{3} =\frac{9}{4} \sin ^{-1} \frac{2\sqrt{2} }{3} (NCERT Miscellaneous Exercise Q12)
12. \tan \left(\frac{\pi }{4} +\frac{1}{2} \cos ^{-1} \frac{a}{b} \right)+\tan \left(\frac{\pi }{4} -\frac{1}{2} \cos ^{-1} \frac{a}{b} \right)=\frac{2b}{a}
13. \sin [\cot ^{-1} \{ \cos (\tan ^{-1} x)\} ]=\frac{\sqrt{x^{2} +1} }{\sqrt{x^{2} +2} }
14. \tan ^{-1} \left(\frac{x}{\sqrt{a^{2} -x^{2} } } \right)=\sin ^{-1} \frac{x}{a} =\cot ^{-1} \left(\frac{\sqrt{a^{2} -x^{2} } }{a} \right)
15. \tan ^{-1} \left(\frac{m}{n} \right)-\tan ^{-1} \left(\frac{m-n}{m+n} \right)=\frac{\pi }{4}
16. \tan ^{-1} \frac{1}{5} +\tan ^{-1} \frac{1}{7} +\tan ^{-1} \frac{1}{3} +\tan ^{-1} \frac{1}{8} =\frac{\pi }{4} (NCERT Miscellaneous Exercise Q8)
17. 4\tan ^{-1} \frac{1}{5} -\tan ^{-1} \frac{1}{70} +\tan ^{-1} \frac{1}{99} =\frac{\pi }{4}
18. \tan ^{-1} \left(\frac{\cos x}{1-\sin x} \right)-\cot ^{-1} \left(\sqrt{\frac{1+\cos x}{1-\cos x} } \right)=\frac{\pi }{4}
19. \sin ^{-1} \frac{3}{5} -\sin ^{-1} \frac{8}{17} =\cos ^{-1} \frac{84}{85} (NCERT Example 10)
20. \sin ^{-1} \frac{8}{17} +\sin ^{-1} \frac{3}{5} =\tan ^{-1} \frac{77}{36} (NCERT Miscellaneous Exercise Q4)

PART - 3

Inverse Trigonometric Functions Lecture 4 Part 3
Play Video
Inverse Trigonometric Functions Lecture 4 Part 3
Play Video

Questions discussed in this lecture

Prove:
21. \tan ^{-1} \frac{63}{16} =\sin ^{-1} \frac{5}{13} +\cos ^{-1} \frac{3}{5} (NCERT Miscellaneous Exercise Q7)
22. \cos ^{-1} \frac{4}{5} +\cos ^{-1} \frac{12}{13} =\cos ^{-1} \frac{33}{65} (NCERT Miscellaneous Exercise Q5)
23. \cos ^{-1} \frac{12}{13} +\sin ^{-1} \frac{3}{5} =\sin ^{-1} \frac{56}{65} (NCERT Miscellaneous Exercise Q6)
24. \sin ^{-1} \frac{12}{13} +\cos ^{-1} \frac{4}{5} +\tan ^{-1} \frac{63}{16} =\pi (NCERT Example 11)
25. 2\tan ^{-1} \left\{\tan \frac{\alpha }{2} \tan \left(\frac{\pi }{4} -\frac{\beta }{2} \right)\right\}=\tan ^{-1} \frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta }
26. 2\tan ^{-1} \left(\sqrt{\frac{a-b}{a+b} } \tan \frac{\theta }{2} \right)=\cos ^{-1} \left(\frac{a\cos \theta +b}{a+b\cos \theta } \right)
27. If \sin \left(\sin ^{-1} \frac{1}{5} +\cos ^{-1} x\right)=1, then find the value of x. (NCERT Exercise 2.2 Q14)
28. If y=\cot ^{-1} (\sqrt{\cos x} )-\tan ^{-1} (\sqrt{\cos x} ), prove that \sin y=\tan ^{2} \frac{x}{2}

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