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

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

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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