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 3 Part 1
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Questions discussed in this lecture

Simplify:
1. \tan ^{-1} \left(\sqrt{\frac{1-\cos x}{1+\cos x} } \right),\, \, x<\pi (NCERT Exercise 2.2 Q7)
2. \tan ^{-1} \left(\frac{\cos x-\sin x}{\cos x+\sin x} \right),\, x<\pi (NCERT Exercise 2.2 Q8)
3. \cot ^{-1} \left(\frac{1}{\sqrt{x^{2} -1} } \right),\, \, |x|\, >1 (NCERT Example 6)
4. \tan ^{-1} \frac{1}{\sqrt{x^{2} -1} } ,\, |x|\, >1 (NCERT Exercise 2.2 Q6)
5. \tan ^{-1} \left(\frac{\sqrt{1+x^{2} } -1}{x} \right),\, \, x\ne 0 (NCERT Exercise 2.2 Q5)

PART - 2

Inverse Trigonometric Functions Lecture 3 Part 2
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Questions discussed in this lecture

Simplify:
6. \tan ^{-1} \frac{x}{\sqrt{a^{2} -x^{2} } } ,\, \, |x|<a (NCERT Exercise 2.2 Q9)

7. \tan ^{-1} \left(\frac{3a^{2} x-x^{3} }{a^{3} -3ax^{2} } \right),\, a>0;\, \frac{-a}{\sqrt{3} } \le x\le \frac{a}{\sqrt{3} } (NCERT Exercise 2.2 Q10)
8. \tan ^{-1} \left(\frac{\cos x}{1+\sin x} \right)
9. \tan ^{-1} \left(\frac{\sin x}{1+\cos x} \right)
10. \tan ^{-1} \left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)
11. \tan ^{-1} \left(\frac{x}{a+\sqrt{a^{2} -x^{2} } } \right)
12. \sin ^{-1} \left(\frac{5}{13} \cos x+\frac{12}{13} \sin x\right)
13. \sin ^{-1} (x\sqrt{1-x} -\sqrt{x} \sqrt{1-x^{2} } )
14. \sin ^{-1} \left\{\frac{\sqrt{1+x} +\sqrt{1-x} }{2} \right\}

PART - 3

Inverse Trigonometric Functions Lecture 3 Part 3
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Questions discussed in this lecture

Simplify:
15. \tan ^{-1} \sqrt{\frac{a-x}{a+x} }
16. \tan ^{-1} \left(\frac{\cos x}{1-\sin x} \right),\, \, -\frac{\pi }{2} <x<\frac{\pi }{2} (NCERT Example 5)
17. \sin ^{-1} \left\{\frac{x+\sqrt{1-x^{2} } }{\sqrt{2} } \right\}
18. \tan ^{-1} (x+\sqrt{1+x^{2} } )
19. \sin \left\{2\tan ^{-1} \sqrt{\frac{1-x}{1+x} } \right\}
20. \sin ^{-1} \left(\frac{\sin x+\cos x}{\sqrt{2} } \right)
21. \tan ^{-1} \left[\frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right],\, {\rm if}\, \, \frac{a}{b} \tan x>-1 (NCERT Example 12)
22. \cot ^{-1} \left(\frac{\sqrt{1+\sin x} +\sqrt{1-\sin x} }{\sqrt{1+\sin x} -\sqrt{1-\sin x} } \right)=\frac{x}{2} ,\, \, x\in \left(0,\, \, \frac{\pi }{4} \right) (NCERT Miscellaneous Exercise Q10)
23. \tan ^{-1} \left(\frac{\sqrt{1+x} -\sqrt{1-x} }{\sqrt{1+x} +\sqrt{1-x} } \right)=\frac{\pi }{4} -\frac{1}{2} \cos ^{-1} x,\, \, -\frac{1}{\sqrt{2} } \le x\le 1 (NCERT Miscellaneous Exercise Q11)

वो खुद ही तय करतें हैं मंज़िल आसमानों की,
परिन्दों को नहीं दी जाती तालीम उड़ानों की,
रखते हैं जो हौंसला आसमां छूने का,
उनको नहीं परवाह कभी गिर जाने की |