INVERSE TRIGONOMETERIC FUNCTIONS

FORMULAE 1- sin-1(sinθ) = θ , cos-1(cos θ) = θ, etc.

2-(i) sin-1(-x) = - sin-1x (ii) cos-1(-x) = π- cos-1x (iii) tan-1(-x) = - tan-1x

(iv) sin-1x = cosec-11/x (v) cos-1x = sec-11/x (v1) tan-1x = cot-11/x

**3Conversion property

(i) sin-1x = cos-1√(1-x2) = tan-1 x/ √(1-x2) (ii) cos-1x = sin-1√(1-x2) = tan-1√(1-x2) / x

(iii) sin-1x+ cos-1x = /2 (iii) tan-1(x) + cot-1x = /2 (iv) cosec-1x +sec-1x = /2

(iv) cosec-11/x = sec-1 1/√(1-x2) = cot-1√(1-x2) / x

***4(Learn)

(i) tan-1x+ tan-1y = tan-1 , xy<1

(ii) tan-1x+ tan-1y = π+tan-1 , xy>1, x>0 , y>0

(iii) tan-1x+ tan-1y= - π+tan-1 , xy>1, x<0 , y<0

(iv) tan-1x+ tan-1y+tan-1z = tan-1

(v) tan-1x- tan-1y = tan-1 , xy > -1, x> 0 , y>0

(vi) tan-1x- tan-1y = π+tan-1 , xy<-1,x>0,y<0

(vii) tan-1x- tan-1y = - π + tan-1 , xy<-1,x<0,y>0

5 Master formula

2tan-1x = tan-1 = sin-1 = cos-1

6 Additional formulae (A) (i) sin-1x+ sin-1y = sin-1 [x √(1-y2) + y√(1-x2)] ,

(II) sin-1x+ sin-1y = - sin-1 [x √(1-y2) + y√(1-x2)] , 0<x, y 1, x2+y2 >1

(III) sin-1x+ sin-1y = - - sin-1 [x √(1-y2) + y√(1-x2)] , -1<x, y<0, x2+y2 >(B) (i) sin-1x- sin-1y = sin-1 [x √(1-y2) - y√(1-x2)] ,

(II) sin-1x- sin-1y = - sin-1 [x √(1-y2) - y√(1-x2)] , 0<x , -1 , x2+y2 >1

(III) sin-1x- sin-1y = - - sin-1 [x √(1-y2) - y√(1-x2)] , -1 x<0 , 0<y , x2+y2 >1 (C) (i) cos-1x + cos-1y = cos-1[xy- ] ,

(II) cos-1x + cos-1y = cos-1[xy- ] , -1 , x+y

1. Express in simplest form

(i) tan-1 ( ) A.

(ii) tan-1 A. x/2

(III) sin [ cot-1{cos(tan-1x)}] A.

(iv) A. x/2

(v) A.sin-1x-sin-1

Q2.Prove that

(i) tan(2tan-11/5 – π /4)= - 7/17

(ii) cot( /4 – 2cot-13) = 7

(iii) tan-11/2 + tan-1 1/3 = π/4

(iv) tan-11+ tan-12 + tan-13 = π

(v) sin-14/5 + cos-12/√5 = cot-12/11

(vi)tan-11/4+tan-12/9=1/2 cos-13/5

(vii) 2sin-13/5 – tan-117/31 = π/4

(viii) tan-11/4 + tan-12/9 = cos-1 2/√5

(ix) sin-1 = tan-1(x2+x+1) (x) 2 tan-1 1/ 5 + sec-15√2 / 7 + 2 tan-11/8 = /4

(xi) sin-1 (1/√10 ) + sin-1 (3/√10) = π/ 2

(xii) cos-1 3/5 + cos-14/5 = / 2

(xiii) tan-1

(xiv) ; then prove x+ y + z = xyz

(xv)

(xvi) 2[ tan-1 ] = (xvii) 2tan-11/3 + tan-11/7 = π/4

(xviii)

(xix)

(xx)

(xxi)

(xxii)

(xxiii) If ,prove that: 9x2 - 12xycosθ + 4y2 = 36sin2θ

(xxiv) If cos-1x+cos-1y+cos-1z=π, prove that:

(xxv) sin-1 3/5 + sin-1 5/13 = sin-1 56/65

Solve:

(i) x=

(ii) x= -1/2

(iii) x=0,-1/2(iv) tan-1(2+x) + tan-1(2-x) = tan-12/3 x=3,-3

(v) x= 1/

(vi) cos-1x + sin-1(x/2) = π/6 x=1(vii) sin-1 x + sin-1 2x= π/3 x=√21/14(viii) tan-12x + tan-13 x= x= -1/6,-1(ix) cos(sin-1x) = 1/9 x= /9(x) tan_!a + cot-1(a+1) = tan-1(a2+a+1)

compiled by – UMENDRA VERMA M.Sc.(Maths)-9415418874