Solution:

Let `cos^(-1) (4/5) =alpha`

`:. cos alpha = 4/5 , sin alpha = 3/5`

`cos^(-1) (12/13) = beta`

`:. cos beta = 12/13 , sin beta = 5/13`

`cos (alpha + beta) = cos alpha cos beta - sin alpha sin beta`

`= 4/5 xx 12/13 - 3/5 xx 5/13 = (48-15)/65 = 33/65`

`:. alpha + beta = cos^(-1) (33/65) `

`:. alpha + beta = cos^(-1) (4/5) + cos^(-1) (12/13) = cos^(-1) (33/65)`

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Solution:

`tan^(-1) (1/5) + tan^(-1) (1/7) + tan^(-1) (1/3) + tan^(-1) (1/8)`

`= tan^(-1) ( ( 1/5+1/7)/( 1-1/5 xx 1/7) ) + tan^(-1) ( (1/3 +1/8)/(1-1/3 xx 1/8) )`

`= tan^(-1) ( (12/35)/(34/35) ) + tan^(-1) ( (11/24)/( 23/24) )`

`=tan^(-1) (12/34) + tan^(-1) (11/23)`

`= tan^(-1) (6/17) + tan^(-1) (11/23)`

`= tan^(-1) ( (6/7 +11/23)/( 1-6/17 xx 11/23) ) = tan^(-1) ( ( (138 +187)/391 )/( (391 -66)/391) )`

`= tan^(-1) (325/325) = tan^(-1) 1 = pi/4`

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Solution:

Putting, `x = tan theta => tan^(-1) x= theta`

`:.` `sin^(-1) ( (2x) /(1+x^2) ) = sin^(-1) ((2 tan theta)/(1+tan^(2) theta))`

`= sin^(-1) (sin 2 theta)`

`= 2 theta = 2 tan^(-1) x`

Put `y = tan phi` `:. phi = tan^(-1) y`


`cos^(-1) ( (1-y^2) /(1+y^2)) = cos^(-1) ((1-tan^(2) phi)/(1+tan^(2) phi))`

`= cos^(-1) (cos 2 phi) ` `\ \ \ \ \ \ \ \ \ \ \ \ \ \ ( :. cos 2 A = (1-tan^(2) A)/(1+tan^(2) A))`

Now ` tan( 1/2 [sin^(-1) ((2x)/(1+x^2)) +cos^(-1) ((1+y^2)/(1-y^2))])`

`= tan( 1/2 [2 tan^(-1) x + 2 tan^(-1) y])`

`= tan (tan^(-1) x +tan^(-1) y)`

`= tan (tan^(-1) ( (x+y)/(1-xy))) = (x+y) /(1-xy)`

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Solution:

Put `x = tan^2 theta , theta = tan^(-1) sqrt (x)`

`R.H.S. = 1/2 cos^(-1) ( (1-x)/(1+x))`

`=1/2 cos^(-1) ((1-tan^2 theta)/( 1+tan^2 theta)) =1/2 cos^(-1) (cos 2 theta)`

`=1/2 xx 2 theta = theta`

`= tan^(-1) sqrt (x)`

` = L.H.S.`

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Solution:

`tan^(-1) ( (sqrt (1+x) +sqrt (1-x) )/( sqrt (1+x) - sqrt (1-x) ) )`

{Put `x = cos 2 theta => theta = 1/2 cos^(-1) x ` }

`= tan^(-1) ( ( sqrt (1+cos 2 theta ) +sqrt (1- cos 2 theta ) )/( sqrt (1+ cos2 theta ) - sqrt ( 1- cos 2 theta) ) )`

`= tan^(-1) ( ( sqrt (2 cos^2 theta ) + sqrt ( 2 sin^2 theta ) )/( sqrt (2 cos^2 theta ) - sqrt (2 sin^2 theta) ) )`

` [ :. 1+ cos 2 theta = 2 cos^2 theta ` and `1-cos 2 theta= 2 sin^2 theta ]`

`= tan^(-1) ( (cos theta + sin theta )/( cos theta - sin theta ) )`

`= tan^(-1) tan ( pi/4 + theta) = pi/4 + theta = pi/4 +1/2 cos^(-1) x`

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Solution:

` (9 pi)/8 - 9/4 sin^(-1) (1/3) = 9/4 sin^(-1) ((2 sqrt (2) )/3)`

`=> 9/4 (sin^(-1) ( (2 sqrt (2) )/3) + sin^(-1) (1/3) ) = (9 pi )/8`

But `sin^(-1) x + sin^(-1) y`

`= sin^(-1) (x sqrt (1-y^2) + y sqrt (1-x^2) )`

`L.H.S. = 9/4 ( sin^(-1) ( (2 sqrt (2) )/3) + sin^(-1) (1/3) )`

`= 9/4 sin^(-1) [ (2 sqrt (2) )/3 sqrt(1-1/9) +1/3 sqrt (1-8/9) ]`

`= 9/4 sin^(-1) [ (2 sqrt (2) )/3 xx (2 sqrt (2) )/3 +1/3 xx 1/3 ]`

` = 9/4 sin^(-1) [ 8/9 +1/9] = 9/4 sin^(-1) 1 = 9/4 xx pi/2 = (9 pi )/8`

Hence `(9 pi)/8 - 9/4 sin^(-1) (1/3) = 9/4 sin^(-1) ( (2 sqrt (2) )/3 )`

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Solution:

`1/2 tan^(-1) x = tan^(-1) \ \ (1-x)/(1+x)`

or `tan^(-1)x = 2 tan^(-1) \ \ (1-x)/(1+x)`


`= tan^(-1) [ ( 2( (1-x)/(1+x) ) )/( 1- ( (1-x)/(1+x))^2 ) ]`

`= tan^(-1) [ (2 (1-x) )/( 1-x) xx ( (1+x)^2 )/( (1+x)^2 - (1-x)^2 )]`

`= tan^(-1) [ (2 (1-x) (1+x) )/(4x) ] = tan^(-1) \ \ (1-x^2 )/(2x)`

`=> x = (1-x^2)/(2x)`

or `2x^2 =1-x^2` or `3x^2 =1 ` or `x = pm 1/sqrt(3)`

But `x ne 1/sqrt(3) ` `:. x = 1/sqrt(3)`

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Solution:

Let `tan^(-1) x = alpha :. tan alpha =x, sin alpha = x/(sqrt (1+x^2) )`

or `alpha = sin^(-1) ( x/( sqrt (1+x^2) ) )`

Now, `sin tan^(-1) x = sin alpha`

`= sin (sin^(-1) (x/(sqrt (1+x^2) )) ) = x/(sqrt (1+x^2) )`

Option (d) is correct.

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Solution:

`tan^(-1) (x/y) - tan^(-1) ( (x-y)/(x+y) )`

`= tan^(-1) [ ( x/y- (x-y)/(x+y) )/( 1+ x/y xx (x-y)/(x+y) ) ]`

` [ :. tan^(-1) a - tan^(-1) b = tan^(-1) ( (a-b)/(1+ab) ) ]`

`= tan^(-1) [ (x (x+y) -y (x-y) )/(y (x+y) +x (x-y) ) ]`

`= tan^(-1) ( (x^2 +xy -xy + y^2 )/( xy + y^2 + x^2 -xy ) )`

`= tan^(-1) ( (x^2 +y^2 )/( x^2 + y^2) ) = tan^(-1) (1)`

`= tan^(-1) (tan pi/4)`

`= pi/4`

`:.` Option (c) is correct.

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Solution:

We know : `2 tan^(-1) x = tan^(-1) \ \ (2x)/(1 - x^2)`

`=> 2 tan^(-1)\ \ 1/5 = tan^(-1) \ \ ( 2 (1/5))/( 1 - (1/5)^2 ) = tan^(-1) \ \ ( 2/5)/((24)/(25)) = tan^(-1) \ \ 5/(12)`

` :. tan ( 2 tan^(-1) \ \ (1/5)) = tan (tan^(-1) \ \ 5/(12) ) = 5/(12)`

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Solution:

To prove `: sin^(-1) (4/5) + sin^(-1) (5/(13)) + sin^(-1) ((16)/(65) ) = pi/2`

Let ` sin^(-1) (4/5) = x`

` => sin x = 4/5`

` => cos x = sqrt(1 - sin^2 x) = 3/5`

` sin^(-1) (5/(13)) = y`

`=> sin y = 5/(13)`

` => cos y = sqrt(1 - sin^2 y) = (12)/(13)`

` sin^(-1) ((16)/(65)) = z`

`=> sin z = (16)/(65)`

`=> cos z = sqrt(1 - sin^2 z) = (63)/(65)`

`tan x = 4/3 , tan y = 5/(13) , tan z = (16)/(63)`

` tan z = (16)/(63) => cot z = (63)/(16)` ...(1)

`tan(x+y) = (tan (x + y))/(1 - tan x . tan y)`

` => tan(x+y) = (4/3 + 5/(13))/(1 - (20)/(36))`

` => tan(x+y) = (63)/(16)`

` => tan(x+y) = cot z ...` [from equation (1)]

` => tan(x+y) = tan (pi/2 - z)`

`=> x + y = pi/2 - z`

`=> x + y + z = pi/2`

`:. sin^(-1) (4/5) + sin^(-1) ( 5/(13)) + sin^(-1) ( (16)/(65)) = pi/2`

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Solution:

` cot^(-1) [ (sqrt(1 + sinx) + sqrt(1 - sinx))/(sqrt(1 + sinx) - sqrt(1 - sinx)) ] `

`= cot^(-1) [ ( sqrt ( sin^2 (x/2) + cos^2 (x/2) + sin 2 (x/2) ) + sqrt ( sin^2 (x/2) + cos^2 (x/2) 1 sin^2 (x/2) ))/(sqrt ( sin 2 (x/2) + cos^2 (x/2) + sin 2 (x/2) ) - sqrt ( sin^2 (x/2) + cos^2 (x/2) - sin 2 (x/2) ))]``

` [Since, `sin^2 A + cos^2 A = 1`]

` = cot^(-1) [ ( sqrt(sin^2 (x/2) + cos^2(x/2) + 2 sin(x/2) cos (x/2)) + sqrt(sin^2 (x/2) + cos^2(x/2) - 2 sin(x/2) cos (x/2)))/(sqrt(sin^2 (x/2) + cos^2(x/2) + 2 sin(x/2) cos (x/2)) - sqrt(sin^2 (x/2) + cos^2(x/2) - 2 sin(x/2) cos (x/2)))]`

[Since,` sin 2A = 2 sinA cosA`]

` = cot^(-1) [ (sqrt(cosx/2 + sin x/2)^2 + sqrt(cosx/2 - sin x/2)^2)/(sqrt(cosx/2 + sin x/2) - sqrt(cosx/2 - sin x/2)^2 ) ]`

` = cot^(-1) ( (2 cos x/2)/(2 sin x/2) ) = cot^(-1) ( cot x/2)`

` = x/2`

Hence proved.

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Solution:

Let `t = tan^(−1) sqrtx`

So `sqrtx = tan t`

i.e. `tan^2 t = x`

On substituting x in the R.H.S. of equation ` tan^(-1) sqrtx = 1/2 cos^(-1) ((1-x)/(1 +x))`,

we get ` 1/2 cos^(-1) ((1-x)/(1 +x)) = 1/2 cos^(-1) ( (1- tan^2 t)/(1+ tan^2 t))`

Now, using the formula ` cos 2 theta = ( (1- tan^2 theta)/(1+ tan^2 theta))` we have

` 1/2 cos^(-1) ((1-x)/(1 +x)) = 1/2 cos^(-1) (cos(2t))`

` = t = tan^(-1) sqrtx = LHS`

Hence Proved.

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Solution:

`tan^(-1) ( (cos x)/(1 + sin x) )`

` = tan^(-1) [ ( sin ( pi/2 -x))/( 1 + cos ( pi/2 -x))]`

` = tan^(-1) [ ( 2 sin ( pi/4 - x/2) cos (pi/4 - x/2))/( 2 cos^2 (pi/4 - x/2))]`

` = tan^(-1) [ tan ( pi/4 - x/2) ] = ( pi/4 - x/2)` (proved)

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Solution:

We know : `2 tan^(-1) x = tan^(-1) \ \ (2x)/(1 - x^2)`

`=> 2 tan^(-1)\ \ 1/5 = tan^(-1) \ \ ( 2 (1/5))/( 1 - (1/5)^2 ) = tan^(-1) \ \ ( 2/5)/((24)/(25)) = tan^(-1) \ \ 5/(12)`

` :. tan ( 2 tan^(-1) \ \ (1/5)) = tan (tan^(-1) \ \ 5/(12) ) = 5/(12)`

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Solution:

`sin^(-1) (1 – x) – 2sin^(-1) x = pi/2`

` => sin^(-1) (1 – x) = pi/2 + 2 sin^(-1) x`

`=> (1 - x) = sin ( pi/2 + 2 sin^(-1) x )`

` => (1 - x) = cos (2 sin ^(-1) x)`

` => (1 - x) = cos ( cos ^(-1) ( 1 - 2x^2 ))`

` => (1 - x) = ( 1 - 2x^2 )`

` => 1 - x = 1 - 2x^2`

`=> 2x^2 - x = 0`

` :. x = 0 , x = 1/2 `

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Solution:

Given that `tan^(-1) ( x _ 1) + tan^(-1) x + tan^(-1) ( x + 1) = tan^(-1) 3x`

`=> tan^(-1) (x -1) + tan^(-1) ( x + 1) = tan^(-1) 3x - tan^(-1) x` ......(1)

We know that, ` tan^(-1) A + tan^(-1) B = tan^(-1) ((A + B)/(1 - AB))`

and `tan^(-1) A - tan^(-1) B = tan^(-1) ((A - B)/(1 + AB))`

Thus, ` tan^(-1) ( x - 1) + tan^(-1) (x + 1) = tan^(-1) ( (x + 1 + x + 1)/( 1 - (x -1)(x + 1)))`

` = tan^(-1) ((2x)/(1 - (x^2 - 1)))`

` = tan^(-1) ( (2x)/(2 - x^2))` ......(2)

Similarly, ` tan^(-1) 3x - tan^(-1) x = tan^(-1) ((3x -x)/(1 + 3x (x)))`

`= tan^(-1) ( (2x)/(1 + 3x^2))`..............(3)

From equations (1), (2) and (3), we have,

` tan^(-1) ( (2x)/(2 - x^2)) = tan^(-1) ( (2x)/(1 + 3 x^2)) `

`=> (2x)/(2 - x^2) = (2x)/(1 + 3 x^2)`

` => 1/(2 - x^2) = (2x)/(1 + 3x^2)`

` => 2 - x^2 = 1 + 3x^2`

` => 4x^2 = 1`

`=> x^2 = 1/4`

` => x = pm 1/2`

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