07 Inverse Trigonometric Functions

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All the formulae required for advanced trigonometric calculations

Transcript of 07 Inverse Trigonometric Functions

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    Inverse Trigonometric Function Key points :

    1. If sin = x, we write = sin-1 x.

    2. sin(sin-1 x) = x, sin-1 (sin) = if

    2,

    2

    cos(cos-1 x)=x, cos-1 (cos) = if [ 0, ]

    tan (tan-1x ) = x, tan-1 (tan) = if

    2,

    2

    3. That value of sin-1 x lying between 2

    and2

    is called the principal value of sin-1 x.

    That value of cos-1 x lying between 0 and is called the principal value of cos-1 x.

    That value of tan-1 x lying between 2

    and2

    is called the principal value of tan-1 x.

    4. If ,1x1 then i) ( ) xsinxsin 11 = ii) cos-1(-x) = xcos 1 5. If Rx , then i) tan-1 (-x) = -tan-1x ii) cot- (-x) = xcot 1 6. If x -1 or x 1, then i) cosec-1 (-x) = -cosec-1x ii) sec-1 (-x) = xsec 1

    7. cosec-1 x = sin-1x

    1 (if x 0).

    sec-1 x = cos-1

    x

    1 (if x 0).

    cot-1 x = tan-1x

    1 (if x > 0).

    = + tan-1x

    1 (if x < 0).

    8. sin-1 x + cos-1 x = /2 , tan-1 x + cot-1 x = /2, sec-1 x + cosec-1 x = /2. 9. If sin-1 x + sin-1 y = /2, then x2 + y2 = 1.

    10. sin(cos-1 x) = 2x1 , cos(sin-1 x) = 2x1 11. 211 x1cosxsin = for x0 1

    = 21 x1cos for 1 x < 0

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    12. 211 x1sinxcos = for o x 1

    = 21 x1sin for -1 x

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    18. ( )21

    x21forx4x3sinxsin3 311 =

    ( ) 1x21forx3x4cosxcos3 311 =

    31

    x31for

    x31xx3

    tanxtan3 23

    11

    =

    19. (a) If -1 < x < 1, -1 < y < 1 and xy< 1, then tan-1 x + tan-1 y = tan-1xy1yx

    +

    (b) If x > 0, y > 0 and xy> 1, then tan-1 x + tan-1 y = tan-1xy1yx

    + +

    (c) If x < 0, y < 0 and xy> 1, then tan-1 x + tan-1 y = tan-1xy1yx

    + -

    20. (a) If xy> -1, then tan-1 x - tan-1 y = tan-1xy1yx

    +

    (b) If x > 0, y < 0 and xy< -1, then tan-1 x - tan-1 y = tan-1 ++

    xy1yx

    (c) If x < 0, y>0 and xy< -1, then tan-1 x - tan-1 y = tan-1xy1yx

    +

    -

    21. tan-1 nm

    + tan-14nm

    nm =

    +

    Or 43

    22. tan-1 nm

    tan-14nm

    nm =

    +

    Or 43

    23. sin(2 tan-1 x ) = 2x1

    x2+

    cos(2 tan-1 x) = 2

    2

    x1x1

    +

    tan(2 tan-1 x ) = 2x1

    x2

    24. sin (3 sin-1 x ) = 3x - 4x3 cos( 3cos-1 x) = 4x3 - 3x

    tan ( 3 tan-1 x) = 23

    x31xx3

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    25. sin ( 4 tan-1x ) = ( )222

    x1

    )x1(x4+

    cos ( 4 tan-1 x ) = ( )2242

    x1

    xx61

    +

    +

    PROBLEMS Very short Answer Questions:

    1. Evaluate the following

    (i) 1 3sin2

    (ii) 1 1cos

    2

    (iii) ( )1sec 2 (iv) ( )1cot 3 Solution:

    (i) { }1 1 1 13 3sin sin sin ( ) sin2 2 3 x x

    = = =

    Solution :

    (ii) 1 1cos42

    =

    Solution :

    (iii) ( )1 1 3sec 2 sec 2 4 4 = = = Solution :

    (iv) ( )1 1 5cot 3 cot 3 6 6 = = = 2. Evaluate the following

    (i) 1 1sin sin3 2

    (ii) 1 3sin sin

    2 2

    (iii) 1 5sin sin6

    (iv) 1 5cos cos4

    Solution :

    (i) 1 11 1sin sin sin sin sin sin 13 2 3 2 3 6 2

    = + = + = =

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    Solution :

    (ii) 1 1 03 3 3sin sin sin sin sin sin1202 2 2 2 2 6 2

    = + = + = =

    Solution :

    (iii) 1 15 1 5 1sin sin sin sin6 2 6 2

    = =

    = 1

    sin6 6 2

    =

    Solution :

    (iv) 1 1 15 1 1 3cos cos cos cos4 4 42 2

    = = = =

    3. Find the values of (i) 1 3sin cos2

    (ii) 1 65tan cos63

    ec

    (iii) 1 4sin 2sin5

    Solution :

    (i) 1 3sin cos5

    1 3 3cos cos5 5

    let = =

    1 3sin cos5

    = 4

    sin5

    =

    3cos

    5

    =

    Solution :

    (ii) 1 165 65 65tan cos cos cos63 63 63

    ec let ec ec = =

    Solution :

    (iii) 1 14 4 4 3sin 2sin sin sin : cos5 5 5 5

    let = = =

    1 4sin 2sin5

    = 2 4 3 24sin 2sin cos 2

    5 5 25 = = =

    4. Evaluate (i): 1 3tan tan4

    (ii) 1 33sin sin7

    (iii) 1 17cos cos6

    Solution :

    (i) { }1 1 13tan 4 (1)4

    Tan Tan Tan = =

    =

    4

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    Solution :

    (ii) 1 1 133 2 2 2sin sin sin sin 5 sin sin7 7 7 7

    = = + =

    Solution :

    (iii) 1 1 117 5cos cos cos cos 3 cos cos6 6 6 6 6

    = = = =

    Short Answer Questions

    (i) Prove the following 1 1 13 8 36sin sin cos5 17 85

    + =

    (ii) 1 1 13 12 33sin cos cos5 13 65

    + =

    Solution :

    (i) L.H.S 1 13 8sin sin5 17

    +

    Let 1 3sin5

    =

    1 8sin17

    =

    3sin

    5 =

    8sin

    17 =

    4cos

    5 =

    15cos

    17 =

    We know that cos( ) cos cos sin sin + =

    4 15 3 8 365 17 5 17 85

    = =

    1 1 1 136 3 8 36cos sin sin cos85 5 17 85

    + = + =

    Solution :

    (ii) 1 3sin5

    let =

    1 12cos13

    =

    3sin

    5 =

    12cos

    13 =

    4cos

    5

    5sin

    13 =

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    Cole know that ( )cos cos cos sin sin + =

    4 12 3 5 33cos( )

    5 13 5 13 65 + = =

    1 33cos65

    + =

    1 1 13 12 33sin cos cos5 13 65

    + =

    Find the values of (i) 1 13 12sin cos cos5 13

    +

    (ii) 1 13 3sin cos

    5 34Tan +

    (iii) 1 13 5cos sin sin5 13

    +

    Solution :

    (i) Let 1 13 12cos cos5 13

    and = =

    3 12cos cos

    5 13and = =

    4 5sin sin

    5 13and =

    ( )1 13 12sin cos cos sin sin cos cos sin5 13

    + = + = +

    4 12 3 55 13 5 13

    = +

    6365

    =

    Solution :

    (ii) Let 1 13 5sin cos5 34

    and = =

    3 5sin cos

    5 34 = =

    4 3cos sin

    5 34 = =

    3 3 27274 5 20( ) 9 20 91 111

    20 20

    Tan TanTanTan Tan

    +++ = = = =

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    Solution :

    (iii) Let 1 13 5sin sin5 13

    and = =

    3 5sin sin

    5 13and = =

    4 12cos cos

    5 13and =

    1 13 5cos sin sin cos( ) cos cos sin sin5 13

    + = + =

    4 12 3 5 335 13 5 13 65

    = =

    3. Show that 2 1 2 1sec ( 2) cos (cot 2) 10Tan ec + = Solution:

    Let 1 12 cot 2Tan and = = 2 cot 2Tan and = = LHS = 2 2 2 2sec cos 1 1 cot 44 1 4 10ec Tan + = + + + = + + = 4. Show that (i) 1 1 11 1 2 0

    7 13 4Tan Tan Tan + + =

    (ii) 1 1 11 1 12 5 8 4

    Tan Tan Tan + + =

    (iii) 1 1 13 3 84 5 19 4

    Tan Tan Tan + = (iv) 1 1 1 11 1 201cot cot (18)7 8 43

    Tan Tan + = +

    Solution :

    (i) Let 1 1 11 1 27 13 9

    Tan Tan Tan = = =

    1 1 27 13 9

    Tan Tan Tan = =

    1 1 2027 13 91( ) 1 901 91

    91 91

    Tan TanTanTan Tan

    +++ = = = =

    2 2( ) 9 9( ) 041 ( ) 1

    8

    Tan TanTanTan Tan

    +

    + = = =+ + +

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    1 1 11 1 20 07 13 9

    Tan Tan Tan + = + =

    Solution :

    (ii) 1 1 11 1 12 5 8

    Tan Tan Tan + +

    Let 1 1 11 1 12 5 8

    Tan Tan Tan = = =

    1 1 12 5 8

    Tan Tan Tan = = =

    1 172 5( ) 1 91

    10

    Tan +

    + = =

    7 156 9 729 8( ) 17 1 72 72 71

    9 8

    Tan + +

    + + = = =

    1 1 11 1 14 2 5 8 4

    Tan Tan Tan + + = + + =

    Solution :

    (iii) Let 1 1 13 3 84 5 19

    Tan Tan and Tan = = =

    3 3 84 5 19

    Tan Tan and Tan = = =

    ( ) ( )( )3 34 5( ) 279 11

    20

    Tan TanTan Tan

    Tan Tan

    + + ++ = = + =

    +

    27 8 513 8811 19 209 127 8 209 2161

    11 19 209

    += = =

    ++

    4

    + =

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    Solution :

    (iv)Let1 11 1

    7 8Tan Tan = =

    1 17 8

    Tan and Tan = =

    1 115 37 8( ) 11 55 111

    50

    Tan TanTanTan Tan

    +++ = = = =

    LHS 1 13 11cot11 3

    Tan + = =

    Let 1 1201

    cot cot 1843

    = =

    201cot cot 18

    43 = =

    201 18 3618 431cot cot 1 43 43cot( ) 201 201 774cot cot

    43 43

    + = = =++

    3575 11975 3

    = =

    1 11cot3

    + = LHS=RHS

    5.Find the value of 1 14 2cos tan5 3

    Tan +

    Solution:

    Let 1 14 2

    cos tan5 3

    and = =

    4 2cos tan

    5 3and = =

    3 2tan

    4 3Tan and = =

    1 1

    3 2 174 2 174 3 12cos tan ( ) 3 2 65 3 1 tan tan 61

    4 3 12

    Tan TanTan Tan

    ++ + = + = = = =

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    6.Prove that 1 11 1 2cos cos4 2 4 2

    a a bTan Tanb b a

    + + =

    Solution:

    Let 1cos cosa a

    b b = =

    L.H.S tan tan4 2 4 2

    + +

    2 2

    2

    1 tan 1 tan1 tan 1 tan 2 22 21 tan 1 tan 1 tan

    2 2 2

    + + +

    + = +

    2

    2

    2 1 tan22 2sec

    1 tan2

    ba

    +

    = =

    7. Solve (i) ( )1 1cos 2sec 9x = (ii) 1 1 13 4

    cos sin cos5 5

    x =

    (iii) ( )1 1 1sin 1 sin cosx x x + = (iv) 1 11sin sin cos 15

    x

    + =

    Solution: Let 1sin sinx x = =

    ( )1 2 21 1 1cos 2sin cos 1 2sin9 9 9x = = =

    2 21 81 2 29 9

    x x = =

    2 4 29 3

    x x= =

    Verification: For 12 2 2sin sin3 3 3

    x = = =

    1 2 22 4 1cos 2sin cos 1 2sin 1 23 9 9

    = = = =

    For 23

    x =

    21 22 2 1cos 2sin 1 2sin 1 2

    3 3 9 = = =

    The Equation gets satisfied for 2 2,

    3 3x =

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    Solution :

    (ii) 1 1 13 4cos sin cos5 5

    x =

    1 3cos5

    Let =

    1 4sin5

    =

    3cos

    5 =

    4sin

    5 =

    Given ( )1cos cosx x = = 3 3 4 4

    cos cos sin sin 15 5 5 5

    x = + = + = 1x =

    Solution :

    (iii) ( )1 1 1sin 1 sin cosx x x + = ( ) ( )11 cos 2sinx x = ( )1 1 1sin 1 cos sinx x x = 1sin sinLet x x = =

    1 1cos sin2

    but x x =

    21 cosx =

    ( )1 1 1sin 1 sin sin2

    x x x

    =

    2 21 1 2sin 1 1 2x x x = =

    ( ) 11 sin 2sin2

    x x

    =

    2 12 0 02

    x x x = = =

    Solution :

    (iv) 1 21sin sin cos5

    x

    +

    =1 1 111 sin cos sin (1)

    5x + =

    1 1 1cos sin2 5

    x

    1 11 1 1cos sin sin sin2 5 5 5

    x

    = = =

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    Long Answer Question

    1 Prove that 1 1 13 5 3232sin cos cos5 13 325

    =

    Let 1 13 5sin cos5 13

    = =

    3 5sin cos

    5 13 = =

    4 12cos sin

    5 13 = =

    2 2 224sin 2sin cos cos 1 2sin25

    = = =

    9 71 225 25

    = =

    We know that ( ) 2 2cos 2 cos cos sin sin = +

    7 5 24 12 35 288 32325 13 25 13 325 325

    += + = =

    Solution :

    (ii) 1 14 1sin 2 tan5 3 2

    + =

    Let 1 4sin5

    = 11

    tan3

    =

    4sin

    5 =

    1tan

    3 =

    3cos

    5 = 2 2

    22 tan 2 9 33sin 11 tan 3 10 51

    9

    = = = =+ +

    2 4cos5

    = We know that cos( 2 ) cos cos sin sin + =

    = 3 4 4 3 04 5 5 5

    =

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    22

    + =

    Solution :

    (iii) 1 1 11 1 14 tan tan tan5 99 70 4

    + =

    Let 1 1tan5

    = 11

    tan99

    = 1 1tan70

    =

    1tan

    5 =

    1tan

    99 = 1tan

    70 =

    22 25 552 1 5 24 121

    25

    Tan = = =

    2

    522 2 120124 251 2 1191144

    TanTanTan

    = = =

    ( ) 441 4Tan TanTan

    Tan Tan

    +

    + =

    =

    120 1 11880 119_ 11999119 99 119 99

    120 1 11781 120 116611119 94

    +

    = =

    ( ) ( )11999 1

    4 11661 704 11999 11 tan(4 ) 111661 70

    Tan TanTan

    Tan

    + + = =

    + + + =

    828269 1828269

    =

    44

    + = 1 1 11 1 14 tan tan tan5 99 70 4

    + =

    2. It 1 1 1cos cos cosp q r + + = prove that 2 2 2 2 1p q r pqr+ + + =

    Solution:

    Let 1cos p = 1cos q = 1cos r = cosp = cosq = cosr = Given + + = ( ) ( )cos cos + = cos cos sin sin cos =

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    2 2 2 21 1 1 1pq r p q pq r p q= + + =

    Squaring on both such

    2 2 2 2 2 2 2 2 2 22 1 2 1p q r pqr p q p q p q r pqr+ + = + + + + =

    3. If 2

    1 1 12 2 2

    2 1 2sin cos

    1 1 1p q xTanp q x

    = + +

    then prove that 1p q

    xpq

    =

    +

    Solution: Let p Tan q Tan = =

    21 1 1

    2 2 22 1 2

    sin cos1 1 1

    Tan Tan xTanTan Tan x

    = + +

    ( ) ( )1 2 1 2 1 22sin sin cos cos 1xTanx

    = =

    Let x Tan=

    1 22 21

    TanTanTan

    =

    2 2 2 = tan

    1 1Tan Tan p q

    Tan Tan pqr

    = =

    + +

    4. If a, b, c are distinct non-zero real numbers having the same sign prove that

    1 1 11 1 1cot cot cot 2ab bc ac ora b b c c a

    + +

    + + =

    Solution: Since a, b, c have same sign two cases will arise

    (i) Two of the three number a-b, b-c,c-a, are positive and there is negative (ii) One of the number is positive and the other two are negative

    Case(i) Suppose a-b>0 b-c>0 and c-a

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    1 1 1

    1x yTan Tan x Tan y

    xy

    = +

    Case (ii) Suppose 0 , 0 0a b b c and c a < < >

    1 1 11 1 1cot cot cotab bc aca b b c a c

    + + + + +

    1 1 1( 1) ( 1) ( 1)cot cot cotab bc cab a c b c a

    + + + = + +

    1 1 11 1 1cot cot cotab bc cab a c b c a

    + + +

    = + +

    1 1 12 tan tan tan1 1 1b a c b c a

    ab bc ac

    = + + + +

    1 1 1 1 12 tan tan tan tan tanb a c b a = + + + = 2

    5. If 1 1 1sin sin sinx y z + + = then prove that ` 2 2 21 1 1 2x x y y z z xyz + + =

    Solution:

    Let 1sin x = 1sin y = 1sin z = sin x = sin y = sin z = Given + + = + = ( ) ( )sin sin cos cosand + = = +

    2 2 21 1 1x x y y z z + +

    2 2 2sin 1 sin sin 1 sin sin 1 sin + +

    2 2 21 sin sin sin2

    + + = ( ) ( ) 21 2sin cos sin2 + +

    ( )1 2sin cos 2sin cos2

    +

    ( ) ( )sin cos cos + + 2sin sin sin = 2xyz=

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    6. If (i) 1 1 1Tan x Tan y Tan z + + = then prove that x y z xyz+ + =

    (ii) 1 1 12

    Tan x Tan y Tan z + + = then prove that 1xy yz zx+ + =

    Solution: Let 1 1 1tan tan tanx y z = = = x Tan= y Tan= z Tan=

    Given x + + = + =

    ( ) ( )1Tan TanTan Tan Tan

    Tan Tan

    +

    + = =

    Tan Tan Tan Tan Tan Tan Tan Tan Tan + = + + + Tan Tan Tan = x y z xyz+ + =

    Solution :

    (ii) 2 2

    + + = + =

    tan tan tan tan 1cot

    1 tan tan 1 tan tan tan

    + +

    = =

    tan tan tan tan 1 tan tan + = tan tan tan tan tan tan 1 + + = 1xy yz zx+ + =

    7. If 2 2

    1

    2 2

    1 11 1

    x xTanx x

    =

    + + then prove that 2 2sinx =

    Solution:

    2 2

    2 2

    1 1tan

    1 1x x

    x x

    +

    + +

    2 2

    2 2

    1 1 1tan 1 1

    x x

    x x

    + + =

    +

    Using componendo and dividend we have

    2 2 2 2

    2 2 2 2

    1 tan 1 1 1 11 tan 1 1 1 1

    x x x x

    x x x x

    + + + + +

    =

    + + + +

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    cos sincos sin

    +

    =

    2

    2

    11

    x

    x

    +

    squaring on both sloves

    2 2 2 2

    2 2 2 2cos sin 2cos sin 1 1 sin 2 1cos sin 2cos sin 1 1 sin 2 1

    x x

    x x

    + + + + ++ =

    +

    2 2

    2 2 2 21 sin 2 1 sin 2 1 1 2 21 sin 2 1 sin 2 1 1 2sin 2sin

    x x

    x x

    + + + + = =

    + + + +

    2 sin 2x =

    8. Prove that 1 1cos coscos 2 tan tan tan1 cos cos 2 2

    + =

    +

    1 1. . 2 tan tan tan tan tan tan

    2 2 2 2R H S Let =

    tan tan tan2 2 =

    . . 2R H S = We know that 2

    21

    cos 21

    TanTan

    +

    2 2

    2 2

    1 tan tan2 2cos 2

    1 tan tan2 2

    =

    + =

    2 2

    2 2

    2 2

    2 2

    sin sin2 2

    cos cos2 2

    sin sin2 2

    .

    cos cos2 2

    2 2 2 2

    2 2 2 2

    1 cos 1 cos 1 cos 1 coscos cos sin sin 2 2 2 22 2 2 2

    1 cos 1 cos 1 cos 1 coscos cos sin sin

    2 2 2 2 2 2 2 2

    + +

    = =

    + + + +

    1 cos cos cos cos 1 cos cos cos cos cos coscos 2

    1 cos cos cos cos 1 cos cos cos cos 1 cos cos

    + + + + + = =

    + + + + + +

    1cos cos cos coscos 2 2 cos1 cos 1 cos cos

    + += = + +

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    1 1 cos cos2 cos2 2 1 cos cos

    Tan Tan Tan

    + = +

    Solve the following Equations for x

    (i) 1 11 12 2 4

    x xTan Tanx x

    + + =

    +

    Let 1 11 12 2

    x xTan Tanx x

    + = = +

    1 12 2

    x xTan Tanx x

    += = +

    Given that 14 1

    Tan TanTan Tan

    +

    + = =

    ( )( )

    ( )( ) ( )( )2 22

    2

    1 11 2 1 22 2 1 1

    4 111

    4

    x xx x x xx x

    x xx

    x

    ++

    + + + +

    = =

    +

    2 222 2 1 2 4 3

    3x x x x

    x+ +

    = =

    2 12 12

    x x= =

    ii) 1 1 1 21 1 2

    2 1 4 1Tan Tan Tan

    x x x

    + = + +

    1 1 12 1 2 1

    Tan Tanx x

    = =+ +

    1 1 14 1 4 1

    Tan Tanx x

    = = + +

    12

    2Tanx

    + =

    22

    1Tan Tan

    Tan Tan x

    +

    =

    ( )( )2

    1 122 1 4 1

    112 1 4 1

    x x

    x

    x x

    ++ +

    =

    + +

    2 24 1 2 1 2

    8 6x x

    x x x y x+ + +

    =

    + +

    3 2 26 2 16 12x x x x+ = +

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    3 26 14 12 0x x x =

    { }22 3 7 6 0x x x = { }22 3 9 2 6 0x x x x + = 2 (3 2)( 3) 0x x x+ =

    2 2 2 25 1 1 51 14 4 2 4 4 2x x x x

    = =

    2 23 5 2 14 4 3

    x xx

    = =

    ( ) ( ) ( )1 1 1a a b c b a b c c a b cTan Tan Tanbc ac ab

    + + + + + +

    + + =

    ( ) ( ) 1a a b c b a b c a b cx xybc ac c+ + + + + +

    = = >

    0 : 0 1x y and xy> > > 1 1 1

    1x yTan x Tan y Tan

    xy

    + = +

    ( ) ( )( ) ( )

    1 1

    1

    a a b c b a b ca a b c b a b c bc acTan Tan

    a b cbc acc

    + + + + + + + +

    + = + + +

    ( ) ( )1

    1

    a a b c b a b cbc acTan

    a b cc

    + + + +

    = +

    =

    ( ) ( )( )( )

    1a b c a b cTan

    a bab c

    + + ++

    1 ( )c a b cTanab

    + +

    = +

    1 ( )c a b cTanab

    + +

    =

    ( ) ( ) ( )1 1 1a a b c b a b c c a b cTan Tan Tanbc ac ab

    + + + + + +

    + =

    ( ) ( ) ( )1 1 1a a b c b a b c c a b cTan Tan Tanbc ac ab

    + + + + + +

    + + =

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    (iii) 2

    1 1 12 2 2

    2 1 23sin 4cos 2 tan1 1 1 3

    x x x

    x x x

    + = + +

    Let 1tan tanx x = =

    21 1 1

    2 2 12 tan 1 tan 2 tan3sin 4cos 2 tan

    1 tan 1 tan 1 tan 3

    + = + +

    { } ( ) ( )1 1 13sin sin 2 4cos cos 2 2 tan tan 23 + =

    1 12 tan tan3 6 6 6 3

    x = = = = =

    (iv) 1 1tan cos sin cot2

    arc arcx

    =

    Let 1cosarcx

    i.e 1 1cos cosx x

    = =

    1 1 1cot cot2 2

    = =

    tan sin =

    2

    21 4

    5x

    x

    =

    2 2 22 4 91 15 5 5

    x x xv

    = = = = 35

    xv

    =

    35

    xv

    = does not satisfy the equation

    (v) ( ) ( )1 1 1 1sin 1 2sin sin 1 2sin2 2

    x x x x

    = = +

    ( ) 11 sin 2sin2

    x x

    = +

    ( ) ( )11 cos 2sinx x = Let 1sin sinx x =

    2 21 cos 1 1 2x x x = =

    ( )2 12 0 1 0 02

    x x x x x = = = =

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    (vi) 1 1cos sin2 6x

    x

    + =

    Let 1 1cos sin2x

    x = =

    cos sin2x

    x = =

    6

    + = 32

    =

    ( )sin sin6

    + =

    1sin cos cos sin

    2 + =

    22 11 1

    4 2 2x x

    x x + =

    ( ) 22 22 11 1 4 2 2x xx =

    2 4 4 22 11

    4 4 4 4 2x x x x

    x + = +

    11.If 1 1 1sin sin sinx y z + + = then prove that { }4 4 4 2 2 2 2 2 2 2 2 24 2x y z x y z x y y z z x+ + + = + + Let 1 1 1sin sin sinx y z = = = sin sin sinx y z = = = ( ) ( )cos cos + + = + =

    2 2 21 1 1x y xy z =

    2 2 21 1 1x y xy z =

    ( ) ( )2 2 2 2 2 21 1 1 2 1x y x y z xy z = +

    2 2 2 2 2 2 2 21 1 2 1x y x y x y z xy z + = +

    Squaring on both sides we have

    2 2 2 22 1z x y xy z =

    ( )24 4 4 2 2 2 2 2 2 2 22 2 2 4 1z x y x z x y y z x y z+ + + =

    4 4 4 2 2 2 2 2 2 2 2 24 2 2 2x y z x y z x y y z x z+ + = + +

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    1 1cos cosx ya b

    + = then prove that 2 2

    22

    2cos sinx xy y

    az ab b + =

    Let 1 1cos cosx yanda b

    = =

    cos cosx ya b

    = = =

    Given + = ( )cos cos cos cos sin sin + =

    2 2

    2 21 1 cosxy x yab a b

    =

    2 2

    2 2cos 1 1xy x yab a b

    =

    Squaring on both sides we have

    2 2 2 22

    2 2 2 2

    2 2

    2 2

    cos 1 1

    2cos

    x y x ya b a bx y xya b ab

    + =

    2 2 2 2 2 22

    2 2 2 2 2 2

    2 2

    2 2

    cos 1

    2cos

    x y x y x ya b a b a yx y xya b ab

    + = +

    2 22

    2 22

    cos 1 cosx xy ya ab b

    + =

    2 22

    2 22

    cos 1 sinx xy ya ab b

    + =

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    PROBLEMS FOR PRACTICE

    1. Find the value of 1 11 1cot cot2 3

    +

    2. Prove that 1 1 14 7 117sin sin sin5 25 125

    + =

    3. Prove that 1 1 14 5 16sin sin sin5 13 25 2

    + + =

    4. Prove that 1 1 41cot 9 cos4 4

    ec

    + =

    5. Show that 1 113 2cot sin sin tan17 3

    =

    6. Prove that 1 14 1sin 2 tan5 3 2

    + =

    7. Prove that 1 11 1cos 2 tan sin 4 tan7 3

    =

    8. Solve ( )15 12sin sin 02

    xx x

    + = >

    9. Solve 1 1 13 4sin sin sin5 5x x

    x + =

    10. Solve 1 1sin sin 23

    x x

    + =

    11. If ( ){ }1 1sin 2 cos cot 2 tan 0x = 12. Prove that ( ){ } 21 1 2 1cos tan sin cot 2xx x + = +