XI XII Mathematics 2nd Paper: November 2022

7.9.1. Symmetric Relation of Inverse Trigonometric Functions

1. `\sin^{-1}\left(x\right)=\cos ec^{-1}\left(\frac1x\right)` when `-1\leq x\leq1`.

2. `\cos ec^{-1}\left(x\right)=\sin^{-1}\left(\frac1x\right)` when `x\leq-1\;or\;x\geq1`.

3. `\cos^{-1}\left(x\right)=\sec^{-1}\left(\frac1x\right)` when `-1\leq x\leq1`.

4. `\sec^{-1}\left(x\right)=\cos^{-1}\left(\frac1x\right)` when `x\leq-1\;or\;x\geq1`.

5. `\tan^{-1}\left(x\right)=\cot^{-1}\left(\frac1x\right)`.

6. `\cot^{-1}\left(x\right)=\tan^{-1}\left(\frac1x\right)`.

Proof: 1. Let `\sin\theta=x\Rightarrow\theta=\sin^{-1}x\cdots\cdots\cdots(1)`.

Again, `\sin\theta=x\Rightarrow\frac1{\cos ec\theta}=x\Rightarrow\cos ec\theta=\frac1x\Rightarrow\theta=\cos ec^{-1}\left(\frac1x\right)\Rightarrow\sin^{-1}\left(x\right)=\cos ec^{-1}\left(\frac1x\right)`.

Similarly, we can prove others relations from 2. to 6.

7. `\sin\left(\sin^{-1}x\right)=x`.

8. `\cos\left(\cos^{-1}x\right)=x`

9. `\tan\left(\tan^{-1}x\right)=x`.

10. `\cot\left(\cot^{-1}x\right)=x`.

11. `\sec\left(\sec^{-1}x\right)=x`.

12. `\cosec\left(\cosec^{-1}x\right)=x`.

Proof:

7. Let `\sin x=\theta\Rightarrow x=\sin^{-1}\theta\Rightarrow x=\sin^{-1}\left(\sin x\right)\Rightarrow \sin^{-1}\left(\sin x\right)=x`.

Similarly, we can prove the relations from 8 to 12.

13. `\sin^{-1}\left(\sin x\right)=x`

14. `\cos^{-1}\left(\cos x\right)=x`.

15. `\tan^{-1}\left(\tan x\right)=x`.

16. `\cot^{-1}\left(\cot x\right)=x`.

17. `\sec^{-1}\left(\sec x\right)=x`

18. `\cosec^{-1}\left(\cosec x\right)=x`.

Proof:

13. `\sin x=\theta\Rightarrow x=\sin^{-1}\theta\Rightarrow x=\sin^{-1}\left(\sin x\right)\Rightarrow\sin^{-1}\left(\sin x\right)=x`.

Similarly, we can prove the relations from 14 to 118.

Example and Assessment:
1. `\sin^{-1}\left(\frac1{\sqrt2}\right)`
`=\sin^{-1}\left(\sin\left(\frac{\pi}4\right)\right)`
`=\frac{\pi}4`.
2. `\sin{\cos^{-1}(\frac{\sqrt3}2)}=\sin{\cos^{-1}(\cos(\frac{\pi}6))}=\sin\frac{\pi}6=\frac1\2`.

Try yourself.
3. `\sin^{-1}\left(-\frac1{\sqrt2}\right)=?`

Try yourself.
4. `\tan{\cos^{-1}(\frac1\{\sqrt3})}=?`

Try yourself.
5. `\sin^{-1}(-1)=\sin^{-1}{-sin(\frac\pi\2)}=\sin^{-1}{sin(-\frac\pi\2)}=-\frac\pi\2`.

Try yourself.
6. `\tan[\tan^{-1}{-\frac1{\sqrt3}}]=\tan[\tan^{-1}{-tan(\frac\pi\6)}]`

`=\tan[\tan^{-1}{tan(-frac\pi\6)}]=-frac\pi\6`.

7. `\sin^{-1}(-\cosx)+\sin^{-1}(\cos3x)=-2x`

Solution:

`\sin^{-1}(-\cosx)+\sin^{-1}(\cos3x)`

`=-\sin^{-1}(\cosx)+\sin^{-1}(\cos3x)`

`=-\sin^{-1}(\sin(\frac(\pi)(2)-x))+\sin^{-1}(\cos(\frac(\pi)(2)-3x))`

`=\frac(\pi)(2)-x+\frac(\pi)(2)-3x`

`=-2x`

8. Prove that, `\sec^2(\tan^{-1}2)+\cosec^2(\cot^{-1}3)=15`. SB 2014

9. Prove that, `\sec^2(\tan^{-1}4)+\tan^2(\sec^{-1}3)=25`. MB 2011

10. Prove that, `\sec^2(\cot^{-1}3+\cosec^2(\tan^{-1}2)=2\frac(13)(36)`. MB 2004

11. Prove that, `\cosec^2(\tan^{-1}\frac(1)(2))-3\sec^2(\cot^{-1}\sqrt3)=1`. KUET 2008-09

7.9.2. `\tan^{-1}x` Related Formulae:

Formula 1:

`\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac(x+y)(1-xy))` when `xy<1`.

Example

1. `\tan^{-1}(\frac1\2)+\tan^{-1}(\frac3\2)=?`

Solution:

So, `\tan^{-1}(\frac1\2)+\tan^{-1}(\frac3\2)`

`=tan^{-1}(\frac(\frac1\2+frac3\2)(1-\frac1\2\times\frac3\2))` since `[xy=\frac1\2\times\frac3\2=\frac3\4<1]`

`=tan^{-1}(\frac(2+3)(4-3))`

`=tan^{-1}5`

2. Prove that, `\tan^{-1}(\frac(7)(11))+\tan^{-1}(\frac(1)(7))=\frac(\pi)(4)`.

Solution: Try yourself.

Formula 2:

`\tan^{-1}x+\tan^{-1}y=\frac(\pi)(2)+\tan^{-1}(\frac(x+y)(1-xy))` when `xy>1`

Example

3. `\tan^{-1}3+\tan^{-1}2=?`

Solution:

So, `\tan^{-1}3+\tan^{-1}2`

`=\pi+\tan^{-1}(\frac(3+2)(1-2\times3))` since `xy=3\times2=6>1`

`=\pi+\tan^{-1}(-1)`

`=\pi-\tan^{-1}(1)`

`=\pi-\frac(\pi)4`

`=\frac(3\pi)4`

4. `\tan^{-1}2+\tan^{-1}3=?`

Solution: Try yourself.

Formula 3:

`\tan^{-1}x+\tan^{-1}y=\frac\pi2` when `xy=1`

Example

5. `\tan^{-1}(\frac3\2)+\tan^{-1}(\frac2\3)=?`

Solution:

So, `\tan^{-1}(\frac3\2)+\tan^{-1}(\frac2\3)`

`=\frac\pi\2` Since `[xy=(\frac3\2)\times(\frac2\3)=1]`

6. `\tan^{-1}(\frac5\11)+\tan^{-1}(\frac11\5)=?`

Formula 4:

`\tan^{-1}x-\tan^{-1}y=\tan^{-1}(\frac(x-y)(1+xy))`

Example

7. Show that, `\tan^{-1}(\frac5\6)-\tan^{-1}(\frac49\71)=\tan^{-1}(\frac1\11)`

Solution:

So, `\tan^{-1}(\frac5\6)-\tan^{-1}(\frac49\71)`

`=\tan^{-1}{\frac(\frac(5)(6)-\frac49\71)(1+(\frac5\6)\times(\frac49\71))}`

`=\tan^{-1}{\frac(355-294)(426+245)}`

`=\tan^{-1}(\frac(1)(11))`. Proved.

8. Prove that, `\tan^{-1}(\frac(1)(3))-\tan^{-1}(\frac(1)(5))+\tan^{-1}(\frac(1)(7))=\tan^{-1}(\frac(1)(11))`.

Solution: Try yourself.

9. `\tan^{-1}(\frac2\10)-\tan^{-1}(\frac3\5)=?`

Solution: Try yourself.

10. `\tan^{-1}2-\tan^{-1}3=?`

Solution: Try yourself.

11. `\tan^{-1}3-\tan^{-1}2=?`

Solution: Try yourself.

Formula 5:

`\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}(\frac(x+y+z-xyz)(1-xy-yz-zx))` when `xy+yz+zx<=1`.

Example

12. Show that, `\tan^{-1}(\frac(1)(4))+\tan^{-1}(\frac(1)(5))+\tan^{-1}(\frac(1)(8))=-\tan^{-1}(\frac(91)(2560))`

Solution:

`\tan^{-1}(\frac(1)(4))+\tan^{-1}(\frac(1)(5))+\tan^{-1}(\frac(1)(8))`

`=\tan^{-1}{\frac(\frac1\4+\frac1\5+\frac1\8-(\frac1\4)(\frac1\5)(\frac1\8))(1-(\frac1\4)(\frac1\5)-(\frac1\5)(\frac1\8)-(\frac1\8)(\frac1\4))}`

since `xy+yz+zx=(\frac(1)(4))(\frac(1)(5))+(\frac(1)(5))(\frac(1)(8))+(\frac(1)(8))(\frac(1)(4))=(\frac(1)(20))+(\frac(1)(40))+(\frac(1)(32))=\frac(17)(160)<1`

`=\tan^{-1}{\frac(40+32+20-1)(160-(40)(32)-(32)(20)-(20)(40))}`

`=\tan^{-1}{\frac(91)(160-1280-640-800)}`

`=\tan^{-1}{\frac(91)(-2560)}`

`=-\tan^{-1}(\frac(91)(2560))`

Formula 6:

`\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\frac(\pi)(2)+\tan^{-1}(\frac(x+y+z-xyz)(1-xy-yz-zx))` when `xy+yz+zx>1`

Example

13. `\tan^{-1}5+\tan^{-1}8+\tan^{-1}2=`

Solution:

`\tan^{-1}5+\tan^{-1}8+\tan^{-1}2`

since `xy+yz+zx=(5)(8)+(8)(2)+(2)(5)=66>1`

`=\frac(\pi)(2)+\tan^{-1}{\frac(5+8+2-(5)(8)(2))(1-(5)(8)-(8)(2)-(2)(5))}`

`=\frac(\pi)(2)+\tan^{-1}{\frac(65)(-65)}`

`=\frac(\pi)(2)+\tan^{-1}(-1)`

`=\frac(\pi)(2)-\tan^{-1}(1)`

`=\frac(\pi)(2)-\frac(\pi)(4)`

`=\frac(\pi)(4)`

Formula 7:

`\2tan^{-1}x=\sin^{-1}\frac(2x)\(1+x^2)=\cos^{-1}\frac(1-x^2)\(1+x^2)=\tan^{-1}(\frac(2x)(1-x^2))`

Example

14. `\tan^{-1}(\frac(1)(2))=\sin^{-1}(?)=\coss^{-1}(?)=\tan^{-1}(?)`

Solution:

`\tan^{-1}(\frac(1)(2))`

`=\frac(1)(2)\times 2\tan^{-1}(\frac(1)(2))`

`=\frac(1)(2)\times\sin^{-1}\frac(2(\frac(1)(2)))\(1+(\frac(1)(2))^2)`

`=\frac(1)(2)\times\sin^{-1}\frac(1)(1+(\frac(1)(4))`

`=\frac(1)(2)\times\sin^{-1}\frac(1)(\frac(5)(4))`

`=\frac(1)(2)\sin^{-1}\(\frac(4)(5))`

`\tan^{-1}(\frac(1)(2))`

`=\frac(1)(2)\times 2\tan^{-1}(\frac(1)(2))`

`=\frac(1)(2)\times\cos^{-1}\frac(1-(\frac(1)(2))^2)\(1+(\frac(1)(2))^2)`

`=\frac(1)(2)\times\cos^{-1}\frac(1-\frac(1)(4))(1+\frac(1)(4))`

`=\frac(1)(2)\times\cos^{-1}(\frac(4-1)(4+1))`

`=\frac(1)(2)\cos^{-1}(\frac(3)(5))`

`\tan^{-1}(\frac(1)(2))`

`=\frac(1)(2)\times 2\tan^{-1}(\frac(1)(2))`

`=\frac(1)(2)\times\tan^{-1}\frac(2(\frac(1)(2)))\(1-(\frac(1)(2))^2)`

`=\frac(1)(2)\times\tan^{-1}\frac(1)(1-(\frac(1)(4))`

`=\frac(1)(2)\times\tan^{-1}\frac(1)(\frac(3)(4))`

`=\frac(1)(2)\tan^{-1}\(\frac(4)(3))`

So, `\tan^{-1}(\frac(1)(2))=\frac(1)(2)\sin^{-1}\(\frac(4)(5))=\frac(1)(2)\cos^{-1}(\frac(3)(5))=\frac(1)(2)\tan^{-1}\(\frac(4)(3))`.

15. Show that, `\4tan^{-1}\frac(1)(5)-\tan^{-1}\frac(1)\(239)=\frac(\pi)(4)`

Solution:

So, `\4tan^{-1}\frac(1)(5)-\tan^{-1}\frac(1)\(239)`

`=\2\times2tan^{-1}\frac(1)(5)-\tan^{-1}\frac(1)\(239)`

`=\2\timestan^{-1}\frac(2\times\frac(1)(5))(1-(\frac(1)(5))^2)-\tan^{-1}\frac(1)\(239)`

`=\2\timestan^{-1}\frac(10)(25-1)-\tan^{-1}\frac(1)\(239)`

`=2\tan^{-1}\frac(5)(12)-\tan^{-1}\frac(1)\(239)`

`=\tan^{-1}\frac(2\times\frac(5)(12))(1-(\frac(5)(12))^2)-\tan^{-1}\frac(1)\(239)`

`=\tan^{-1}\frac(\frac(5)(6))(1-\frac(25)(144))-\tan^{-1}\frac(1)\(239)`

`=\tan^{-1}\frac(120)(144-25)-\tan^{-1}\frac(1)\(239)`

`=\tan^{-1}\frac(120)(119)-\tan^{-1}\frac(1)\(239)`

`=\tan^{-1}\frac(\frac(120)(119)-\frac(1)\(239))(1+(\frac(120)(119))(\frac(1)\(239))`

`=\tan^{-1}(\frac(\frac(2856)(28441))(\frac(2856)(28441)))`

`=\tan^{-1}(1)`

`=\frac(\pi)(4)`

Formula 8:

`\2tan^{-1}x=\pi+\tan^{-1}(\frac(2x)(1-x^2))` when `x>1`

Example

16. `2\tan^{-1}(\frac(3)(2))=?`

Solution:

`2\tan^{-1}(\frac(3)(2))`

`=\pi+\tan^{-1}(\frac(2\times\frac(3)(2))(1-(\frac(3)(2))^2))`

`=\pi+\tan^{-1}(\frac(3)(1-(\frac(9)(4))))`

`=\pi+\tan^{-1}(\frac(3)(4-9)`

`=\pi-\tan^{-1}(\frac(3)(5))`

17. Prove that, `2\tan^{-1}(\frac(3)(2))+tan^{-1}(\frac(3)(5))=\pi`

18. Prove that, `\sin^{-1}(\frac(3)(5))+\cot^{-1}(\frac(17)(19))=\tan^{-1}(\frac(127)(11))`

Solution:

`\sin^{-1}(\frac(3)(5))+\cot^{-1}(\frac(17)(19))`

`=\tan^{-1}(\frac(3)(4))+\tan^{-1}(\frac(19)(17))`

`=\tan^{-1}{\frac(\frac(3)(4)+\frac(19)(17))(1-(\frac(3)(4))(\frac(19)(17)))}`

`=\tan^{-1}{\frac(51+76)(68-57)}`

`=\tan^{-1}(\frac(127)(11))`

19. Prove that, `\sin^{-1}(\frac(4)(5))+\cos^{-1}(\frac(2)(\sqrt5))=\tan^{-1}(\frac(11)(2))`

Solution: Try yourself.

20. Prove that, `4{\sin^{-1}(\frac(1)(\sqrt5))+\cot^{-1}3)}=\pi`

Solution: Try yourself.

21. Prove that, `\tan^{-1}(\frac(1)(4))+\tan^{-1}(\frac(2)(9))=\frac(1)(2)\cos^{-1}(\frac(3)(5))`

Solution: Try yourself.

22. Prove that, `\cos^{-1}(\frac(63)(65))+2\tan^{-1}(\frac(1)(5))=\tan^{-1}(\frac(3)(4))`

Solution: Try yourself.

23. Prove that, `\cos^{-1}(\frac(1)(\sqrt5))-\frac(1)(2)\sin^{-1}(\frac(3)(5))+\tan^{-1}(\frac(1)(3))=\tan^{-1}2`

Solution: Try yourself.

24. Prove that, `\sin^{-1}(\frac(3)(5))+\frac(1)(2)\cos^{-1}(\frac(5)(13))-\cot^{-1}2=\tan^{-1}(\frac(28)(29))`

Solution: Try yourself.

25. Prove that, `\tan^{-1}(\frac(2)(3))=\frac(\pi)(2)-\sec^{-1}(\frac(sqrt13)(2)`.

Solution: Try yourself.

26. Prove that, `\cos^{-1}(\sqrt(\frac(2)(3)))-\cos^{-1}(\frac(\sqrt6+1)(2sqrt3))=\frac(\pi)(6)`

Solution: Try yourself.

27. Prove that, `\cos(2\tan^{-1}(\frac(1)(7)))=\frac(\pi)(6)`

Solution: Try yourself.

28. Prove that, `\tan(2\tan^{-1}x)=2\tan(\tan^{-1}x+\tan^{-1}x^3)`. ***

Solution: Try yourself.

29. Prove that, `\cot^{-1}(\tan2x)+\cot^{-1}(-\tan3x)=x`

Solution: Try yourself.

30. Prove that, `\tan^{-1}(\frac(1)(2)\tan2\theta)+\tan^{-1}(\cot\theta)+\tan^{-1}(\cot^3\theta)=0`.*

Solution: Try yourself.

31. Prove that, `\tan^{-1}{(\sqrt2+1)\tan\theta}-tan^{-1}{(\sqrt2-1)tan\theta}=\tan^{-1}(\sin2\theta)`.*

Solution: Try yourself.

32. If , `\sin^{-1}(\frac(2p)(1+p^2))+\cos^{-1}(\frac(1-q^2)(1+q^2))=2\tan^{-1}x` then prove that, `x=\frac(p-q)(1+pq)`.*

Solution: Try yourself.

33. Prove that, `\tan^{-1}(\sqrtx)=\frac(1)(2)cos^{-1}(\frac(1-x)(1+x))`.*

Solution: Try yourself.

34. Prove that, `\tan^{-1}x=\frac(1)(2)cosec^{-1}(\frac(1+x^2)(2x))`.*

Solution: Try yourself.

35. Prove that, `\tan^{-1}x=\frac(1)(2)cosec^{-1}(\frac(1+x^2)(2x))`.*

Solution: Try yourself.

36. Prove that, `2\tan^{-1}(\sqrt(\frac(x)(y))\tan\frac(\theta)(2))=sin^{-1}{\frac(2\sqrt(xy)\sin\theta)((y+x)+(y-x)\cos\theta)}`.*

Solution:

`2\tan^{-1}(\sqrt(\frac(x)(y))\tan\frac(\theta)(2))`

`=sin^{-1}(\frac(2\sqrt(\frac(x)(y))\tan\frac(\theta)(2))(1+(\sqrt(\frac(x)(y))\tan\frac(\theta)(2))^2))`

`=sin^{-1}(\frac(2\sqrt(\frac(x)(y))\tan\frac(\theta)(2))(1+\frac(x)(y)\tan^2\frac(\theta)(2)))`

`=sin^{-1}(\frac(2\sqrt(xy)\frac(\sin\frac(\theta)(2))(\cos\frac(\theta)(2)))(y+x\frac(\sin^2\frac(\theta)(2))(\cos^2\frac(\theta)(2))))`

`=sin^{-1}(\frac(\sqrt(xy)\times2\sin\frac(\theta)(2)\cos\frac(\theta)(2))(y\cos^2\frac(\theta)(2)+x\sin^2\frac(\theta)(2)))`

`=sin^{-1}{\frac(\sqrt(xy)\times\sin\theta)(\frac(y)(2)\times2\cos^2\frac(\theta)(2)+\frac(x)(2)\times2\sin^2\frac(\theta)(2))}`

`=sin^{-1}{\frac(2\sqrt(xy)\times\sin\theta)(y(1+\cos\theta)+x(1-\cos\theta))}`

`=sin^{-1}{\frac(2\sqrt(xy)\sin\theta)((y+x)+(y-x)\cos\theta)}`.

37. Prove that, `2\tan^{-1}(\sqrt(\frac(x-y)(x+y))\tan\frac(\theta)(2))=cos^{-1}(\frac(y+x\cos\theta)(x+y\cos\theta))`.***

Solution: Try yourself.

38. Prove that, `2\tan^{-1}[\tan\frac(\alpha)(2)\tan(\frac(\pi)(4)-\frac(\beta)(2))]=tan^{-1}(\frac(\sin\alpha\cos\beta)(\sin\beta+\cos\alpha))`.*

Solution: Try yourself.

39. Prove that, `\tan^{-1}(\frac(x\cos\theta)(1-x\sin\theta))-\tan^{-1}(\frac(x-\sin\theta)(cos\theta))=\theta`.*

Solution: Try yourself.

7.9.3. Formulae for `\sin^{-1}x+\cos^{-1}x`, `\tan^{-1}x+\cot^{-1}x` and `\cosec^{-1}x+\sec^{-1}x`:

`\sin^{-1}x+\cos^{-1}x=\frac\pi\2` when `-1<=x<=1`

`\tan^{-1}x+\cot^{-1}x=\frac\pi\2` when `x>=0`

`\cosec^{-1}x+\sec^{-1}x=\frac\pi\2` when `x<=-1, x>=1`

3.9.4. Formula for `\sin^{-1}x+-\sin^{-1}y` and `\cos^{-1}x+-\sin^{-1}y`:

`\sin^{-1}x+\sin^{-1}y=\sin^{-1}{x\sqrt(1-y^2)+y\sqrt(1-x^2)}` when `x^2+y^2<=1`

`\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}{x\sqrt(1-y^2)+y\sqrt(1-x^2)}` when `x^2+y^2>1`

`\sin^{-1}x-\sin^{-1}y=\sin^{-1}{x\sqrt(1-y^2)-y\sqrt(1-x^2)}`

`\cos^{-1}x+\cos^{-1}y=\cos^{-1}{xy-\sqrt((1-y^2)(1-y^2))}` when `x+y>=0`

`\cos^{-1}x-\cos^{-1}y=\cos^{-1}{xy+\sqrt((1-y^2)(1-y^2))}`

Example and Assessment

1. If `\sin^{-}x+\sin^{-}\frac(\pi)(2)`, then prove that, (i) `x^2+y^2=1` and (ii) `x\sqrt(1-y^2)+y\sqrt(1-x^2)`.

Try yourself

7.9.5. Formula for `2\sin^{-1}x`, `2\cos^{-1}x`, `2\tan^{-1}x`, `3\sin^{-1}x`, `3\cos^{-1}x` and `3\tan^{-1}x`:

Formula

(1) `2\sin^{-1}x=sin^{-1}(2x\sqrt(1-x^2))`

Proof:

(1) Let `\sin^{-1}x=y\Rightarrow x=\siny-----(i)`

We know that,

`\sin2y=2\siny\cosy=2siny\sqrt(1-\sin^2y)=2x\sqrt(1-x^2)`

`\Rightarrow \sin2y=2x\sqrt(1-x^2)`

`\Rightarrow 2y=\sin^{-1}(2x\sqrt(1-x^2))`

`\Rightarrow 2\sin^{-1}=\sin^{-1}(2x\sqrt(1-x^2))`

Formula (2) `2\cos^{-1}x=cos^{-1}(2x^2-1)`

Proof:

Let `\cos^{-1}x=y\Rightarrow x=\cosy-----(i)`

We know that,

`\cos2y=2\cos^2y-1=2x^2-1`

`\Rightarrow \cos2y=2x^2-1`

`\Rightarrow 2y=\cos^{-1}(2x^2-1)`

`\Rightarrow 2\cos^{-1}=\cos^{-1}(2x^2-1)`

Formula (3) `2\tan^{-1}x=tan^{-1}(\frac(2x)\(1-x^2))`

Proof:

Let `\tan^{-1}x=y\Rightarrow x=\tany-----(i)`

We know that,

`\tan2y=\frac(2\tany)\(1-\tany)`

`\Rightarrow 2y=\tan^{-1}(\frac(2\tany)\(1-\tan^2y))`

`\Rightarrow 2\tan^{-1}x=\tan^{-1}(\frac(2x)\(1-x^2))`

Formula (4) `3\sin^{-1}x=sin^{-1}(3x-4x^3)`

Proof:

Let `\sin^{-1}x=y\Rightarrow x=\siny-----(i)`

We know that,

`\sin3y=3\siny-4sin^3y`

`\Rightarrow 2y=\tan^{-1}(\frac(2\tany)\(1-\tan^2y))`

`\Rightarrow 2\tan^{-1}x=\tan^{-1}(\frac(2x)\(1-x^2))`

Example

(1) Prove that, `\sin(3\sin^{-1}x)=3x-4x^3`.

Solution:

`\sin(3\sin^{-1}x)`

`\sin(sin^{-1}(3x-4x^3))`

`=3x-4x^3`. Proved

Formula (5) `3\cos^{-1}x=cos^{-1}(4x^3-3x)`

Proof:

Formula (6) `3\tan^{-1}x=tan^{-1}(\frac(3x-x^3)\(1-3x^2))`

Proof:

Example and Assessment

(1) Prove that, `\cos^{-1}x=2\sin^{-1}\sqrt(\frac(1-x)(2))=2\cos^{-1}\sqrt(\frac(1+x)(2))`.

RUET 2004-05

Solution:

Let `\cos^{-1}x=\theta\Rightarrowx=\cos\theta`.

Now,

`\sin(\frac(\theta)(2))=\sqrt(\frac(2sin^2\frac(\theta)(2))(2))`

`\Rightarrow\sin(\frac(\theta)(2))=\sqrt(\frac(1-\cos\theta)(2))`

`\Rightarrow\sin(\frac(\theta)(2))=\sqrt(\frac(1-x)(2))`

`\Rightarrow\frac(\theta)(2)=\sin^{-1}{\sqrt(\frac(1-\cos\theta)(2))}`

`\Rightarrow\cos^{-1}x=2\sin^{-1}\sqrt(\frac(1-x)(2))`

Again,

`\cos(\frac(\theta)(2))=\sqrt(\frac(2cos^2\frac(\theta)(2))(2))`

`\Rightarrow\cos(\frac(\theta)(2))=\sqrt(\frac(1+\cos\theta)(2))`

`\Rightarrow\cos(\frac(\theta)(2))=\sqrt(\frac(1+x)(2))`

`\Rightarrow\frac(\theta)(2)=\cos^{-1}{\sqrt(\frac(1+\cos\theta)(2))}`

`\Rightarrow\cos^{-1}x=2\cos^{-1}\sqrt(\frac(1+x)(2))`

So, `\cos^{-1}x=2\sin^{-1}\sqrt(\frac(1-x)(2))=2\cos^{-1}\sqrt(\frac(1+x)(2))`.

The least numeral value (positive or negative) of inverse trigonometric function is known as principal value of inverse trigonometric function.

Inverse Trigonometric Functions	Principal Value when `x\geq0`	Principal Value when `x<0`
`\sin^{-1}x`	`0\leq\sin^{-1}(x)\leq\frac{\pi}2`	`-\frac{\pi}2\leq\sin^{-1}(x)<0`
`\cos^{-1}x`	`0\leq\cos^{-1}(x)\leq\frac{\pi}2`	`\frac{\pi}2\leq\cos^{-1}(x)\leq\pi`
`\tan^{-1}x`	`0\leq\tan^{-1}(x)\leq\frac{\pi}2`	`-\frac{\pi}2\leq\tan^{-1}(x)<0`
`\cot^{-1}x`	`0\leq\cot^{-1}(x)\leq\frac{\pi}2`	`\frac{\pi}2\leq\cot^{-1}(x)\leq\pi`
`\sec^{-1}x`	`0\leq\sec^{-1}(x)\leq\frac{\pi}2`	`\frac{\pi}2\leq\sec^{-1}(x)\leq\pi`
`\cosec^{-1}x`	`0\leq\cosec^{-1}(x)\leq\frac{\pi}2`	`-\frac{\pi}2\leq\cosec^{-1}(x<0`

Example

1. Find the principal value of `\sin^{-1}\left(\frac1\2right)`.

Solution: We know that,

`\sin^{-1}(\frac(1)(2))=\sin^{-1}{\sin(\frac(pi)(6))}=\frac(pi)(6)` and `\sin^{-1}(\frac(1)(2))=\sin^{-1}{\sin(2\pi+\frac(pi)(6))}=\frac(13\pi)(6)`.

Since, `0\leq\sin^{-1}(x)\leq\frac(\pi)(2)` for `x\geq0` and `x=\frac(1)(2)`

So, the principal value of `\sin^{-1}\left(\frac1\2right)` is `\frac\pi\6`. (Answer:)

2. Find the principal value of `\cos^{-1}\left(\frac1{\sqrt2}\right)`.

Solution: We know that,

`\cos^{-1}(\frac(1)(\sqrt2))=\cos^{-1}{(\cos(\frac(\pi)(4)}=\frac(\pi)(4)` and `\cos^{-1}(\frac(1)(\sqrt2))=\cos^{-1}{\cos(2\pi+\frac(pi)(4))}=\frac(9\pi)(4)`.

Since, `0\leq\cos^{-1}(x)\leq\frac(\pi)(2)` for `x\geq0` and `x=\frac(1)(\sqrt2)`

So, the principal value of `\cos^{-1}\left(\frac1{\sqrt2}\right)` is `\frac\pi\4`. Answer:

3. Find the principal value of `\sin^{-1}\left(-\frac1{\sqrt2}\right)`.

Solution: We know that,

`\sin^{-1}\left(-\frac1{\sqrt2}\right)` = `\sin^{-1}\left(-sin\frac\pi\4\right)` = `\sin^{-1}\{sin\left(-\frac\pi\4\right)\}` = `-\frac\pi\4`

Since, `-\frac(\pi)(2)\leq\sin^{-1}(x)\leq0` for `x<0` and `x=-\frac(1)(\sqrt2)`

So, the principal value of `\sin^{-1}\left(-\frac1{\sqrt2}\right)` is `-\frac\pi\4`. Answer:

4. Find the principal value of `\sin^{-1}\left(-1right)`.

Try yourself.

5. Find the principal value of `\sin^{-1}\left(-\frac\sqrt3\2right)`.

Try yourself.

6. Find the principal value of `\sin^{-1}\left(\frac\sqrt3\2right)`.

Try yourself.

7. Find the principal value of `\cos^{-1}\left(\frac\sqrt3\2right)`.

Try yourself.

8. Find the principal value of `sin{\cos^{-1}\left(\frac\sqrt3\2right)}`.

Try yourself.

9. Find the principal value of `tan{\cos^{-1}\left(\frac1{\sqrt2}right)}`.

Try yourself.

10. Find the principal value of `\sin^{-1}\left(\frac1\2right)`.

Try yourself.

11. Find the principal value of `\tan^{-1}8+\tan^{-1}\left(\frac9\7right)`.

Solution: We know that,

So, the principal value of `\tan^{-1}8+\tan^{-1}\left(\frac9\7right)` is `\tan^{-1}\left(\frac(8+\frac9\7)\(1-8*\frac9\7)right)` = `\tan^{-1}\left(\frac(56+9)\(7-72)right)` = `\tan^{-1}(-1)` = `\tan^{-1}(-\tan\frac\pi\4)` = `\tan^{-1}{\tan(-\frac\pi\4)}`= `\tan^{-1}{\tan(-\frac\pi\4)}`, `\tan^{-1}{\tan(\pi-\frac\pi\4)}` = `-\frac\pi\4`, `\frac{3\pi}4` = `\frac{3\pi}4`. [Since sum of two positive angles cannot be negative]