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 5 Part 1
Play Video

Questions discussed in this lecture

Solve:
1. \tan ^{-1} \frac{x-1}{x-2} +\tan ^{-1} \frac{x+1}{x+2} =\frac{\pi }{4} (NCERT Exercise 2.2 Q15)
2. \tan ^{-1} 2x+\tan ^{-1} 3x=\frac{\pi }{4} (NCERT Example 13)
3. \tan ^{-1} (x+1)+\tan ^{-1} (x-1)=\tan ^{-1} \frac{8}{31}
4. 2\tan ^{-1} (\cos x)=\tan ^{-1} (2\, {\rm cosec}\, x) (NCERT Miscellaneous Exercise Q13)
5. \tan ^{-1} \frac{1-x}{1+x} =\frac{1}{2} \tan ^{-1} x,\, (x>0) (NCERT Miscellaneous Exercise Q14)
6. \sin ^{-1} (1-x)-2\sin ^{-1} x=\frac{\pi }{2} (NCERT Miscellaneous Exercise Q16)
7. \sin ^{-1} x+\sin ^{-1} (1-x)=\cos ^{-1} x
8. \sin ^{-1} 6x+\sin ^{-1} 6\sqrt{3} x=-\frac{\pi }{2}

PART - 2

Inverse Trigonometric Functions Lecture 5 Part 2
Play Video

Questions discussed in this lecture

Solve:
9. \tan ^{-1} (x-1)+\tan ^{-1} x+\tan ^{-1} (x+1)=\tan ^{-1} 3x
10. 3\sin ^{-1} \frac{2x}{1+x^{2} } -4\cos ^{-1} \frac{1-x^{2} }{1+x^{2} } +2\tan ^{-1} \frac{2x}{1-x^{2} } =\frac{\pi }{3}
11. If \sin ^{-1} \frac{2a}{1+a^{2} } -\cos ^{-1} \frac{1-b^{2} }{1+b^{2} } =\tan ^{-1} \frac{2x}{1-x^{2} } , then prove that x=\frac{a-b}{1+ab}.
12. Evaluate: \tan ^{-1} \left(\frac{a+bx}{b-ax} \right),\, \, x<\frac{b}{a}
13. Prove: \tan ^{-1} \left(\frac{a-b}{1+ab} \right)+\tan ^{-1} \left(\frac{b-c}{1+bc} \right)+\tan ^{-1} \left(\frac{c-a}{1+ca} \right)=0
14. If \tan ^{-1} x+\tan ^{-1} y=\frac{4\pi }{5} , then find the value of \cot ^{-1} x+\cot ^{-1}y?
15. If \tan ^{-1} \left(\frac{1}{1+1.2} \right)+\tan ^{-1} \left(\frac{1}{1+2.3} \right)+…+\tan ^{-1} \left(\frac{1}{1+n.(n+1)} \right)=\tan ^{-1} \phi , then find the value of \phi.
16. If (\tan ^{-1} x)^{2} +(\cot ^{-1} x)^{2} =\frac{5\pi ^{2} }{8} , then find x.

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