Solution:

We know that `sin^(−1)(sin x) = x` . Therefore, `sin ^(-1) (sin \ (3 pi)/5) = (3 pi)/5`

But `(3 pi)/5 notin [ -pi/2 , pi/2]`, which is the principal branch of `sin^(–1) x`

However `sin ((3 pi)/5)= sin (pi-(3pi)/5)= sin \ (2 pi)/5` and `(2 pi)/5 in [ - pi/2 , pi/2]`

Therefore `sin^(-1) (sin \ (3 pi)/5) = sin^(-1) (sin \ (2 pi)/5) =(2 pi)/5`

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Solution:

Let `sin^(-1) \ 3/5 = x` and `sin ^(-1) \ 8/17 = y`

Therefore `sin x = 3/5` and `sin y = 8/17`

Now `cos x = sqrt(1- sin^2 x)= sqrt(1- 9/25) = 4/5`

and `cos y = sqrt(1- sin^2 y)= sqrt(1- 64/289)= 15/17`

We have `cos (x–y) = cos x cos y + sin x sin y`

`=4/5 xx 15/17 + 3/5 xx 8/17 = 84/85`

Therefore `x-y = cos^(-1) \ (84/85)`

Hence ` sin^(-1) \ 3/5 - sin^(-1) \ 8/17 = cos^(-1) \ 84/85`

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Solution:

Let `sin^(-1) \ 12/13 = x , cos ^(-1) \ 4/5= y, tan^(-1) \ 63/16= z`

Then `sinx= 12/13, cos y= 4/5, tan z= 63/16`

Therefore `cos x= 5/13, sin y= 3/5, tan x= 12/5` and `tan y = 3/4`

We have `tan (x+y)=(tan x+ tan y)/(1- tan x tan y) = (12/5 + 3/4)/(1- 12/5 xx 3/4) = -63/16`

Hence `tan(x + y) = − tan z`

i.e., `tan (x + y) = tan (–z)` or `tan (x + y) = tan (π – z)`

Therefore `x + y = – z` or `x + y = π – z`

Since `x, y` and `z` are positive, `x + y ≠ – z`

Hence `x+y+z= pi` or `sin^(-1) \ 12/13 + cos^(-1) \ 4/5 + tan^(-1) \ 63/16 = pi`

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Solution:

We have,

`tan^(-1) [(a cos x- b sin x)/(b cos x + a sin x)]= tan^(-1) [ ((a sin x- b sin x)/(b cos x))/((b cos x + a sin x)/(b cos x))]= tan^(-1) [ (a/b - tan x)/(1+ a/b tan x)]`

`= tan^(-1) \ a/b - tan^(-1) (tan x) = tan^(-1) \ a/b-x`

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Solution:

We have `tan^(–1) 2x + tan^(–1) 3x= pi/4`

or, `tan^(-1) ((2x+3x)/(1-2x xx 3x))= pi/4`

i.e., `tan^(-1) ((5x)/(1-6x^2))= pi/4`

Therefore `(5x)/(1-6x^2) =tan \ pi/4=1`

or `6x^2 + 5x – 1 = 0` i.e., `(6x – 1) (x + 1) = 0`

which gives `x=1/6` or `x=-1`

Since `x = – 1` does not satisfy the equation, as the L.H.S. of the equation becomes negative, `x =1/6` is the only solution of the given equation.

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Solution:

`cos^(-1) (cos (13 pi)/6) = cos^(-1) cos (2 pi + pi/6)`

`= cos^(-1) cos pi/6 = pi/6`

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Solution:

`tan^(-1) (tan ((7 pi)/6)) = tan^(-1) tan (pi + pi/6)`

`= tan^(-1) tan (pi/6) = pi/6`

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Solution:

Let `sin^(-1) x = alpha ` , `:. sin alpha = x`,

`cos alpha = sqrt (1-x^2) , sin^(-1) y = beta`

`:. sin beta = y, cos beta = sqrt (1- y^2 )`

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

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

`alpha + beta = sin^(-1) [x sqrt (1-y^2) + y sqrt (1-x^2) ]`

i.e., `sin^(-1) x + sin^(-1) y`

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

`:. sin^(-1) 8/17 + sin^(-1) 3/5`

`= sin^(-1) [ 8/17 sqrt (1-9/25) +3/5 sqrt (1- 64/289) ]`

`= sin^(-1) [8/17 xx 4/5 + 3/5 xx 15/17]`

`= sin^(-1) [(32+45)/85] = sin^(-1) 77/85`

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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:

Let `cos^(-1) (12/13) = alpha`,

`:. cos alpha = 12/13 , sin alpha = 5/13, sin^(-1) (3/5) = beta`

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

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

`= 5/13 xx 4/5 +12/13 xx 3/5 = (20 + 36)/65 = 56/65`

`:. alpha + beta = sin^(-1) (56/65)`

`=> cos^(-1) (12/13) + sin^(-1) ( 3/5) = sin^(-1) (56/65)`

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Solution:

Let `cos^(-1) (12/13) = alpha`,

`:. cos alpha = 12/13 , sin alpha = 5/13, sin^(-1) (3/5) = beta`

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

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

`= 5/13 xx 4/5 +12/13 xx 3/5 = (20 + 36)/65 = 56/65`

`:. alpha + beta = sin^(-1) (56/65)`

`=> cos^(-1) (12/13) + sin^(-1) ( 3/5) = sin^(-1) (56/65)`

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Solution:

Let `sin^(-1) (5/13) = alpha , sin alpha = 5/13, tan alpha = 5/12`

`cos^(-1) (3/5) = beta , cos beta = 3/5 , tan beta = 4/3`

`tan (alpha + beta) = (tan alpha + tan beta)/( 1- tan alpha tan beta) = (5/12 + 4/3 )/(1- 5/12 xx 4/3)`

`= (15+48)/( 36- 20) = 63/16`

`:. alpha + beta = tan^(-1) (63/16)`

Thus `tan^(-1) (63/16) = sin^(-1) (5/13) + cos^(-1) (3/5)`

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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:

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:

`cot^(-1) ( (sqrt (1+sin x ) + sqrt (1- sin x) )/( sqrt (1+sin x) - sqrt (1- sin x) ))`

Now, `1+ sin x = cos^2 (x/2) + sin^2 (x/2) + sin^2 (x/2) + 2 cos (x/2) sin (x/2)`

`= (cos (x/2) + sin (x/2) )^2`

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

`:. cot^(-1) ( (sqrt (1+sin x) + sqrt (1- sin x) )/( sqrt (1+ sin x) - sqrt (1- sin x) ) )`

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

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

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

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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:

Now, `L.H.S. = 2 tan^(-1) (cos x)`

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

Putting this value in given equation, we get

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

`=> (2 cos x)/(sin^2 x) = 2 cosec x = 2/(sin x)`

`=> cos x = sin x` or `tan x = 1 => x = pi/4`

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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:

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

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

or `sin^(-1) (1-x) -2 sin^(-1) x = sin^(-1) (1-x) + cos^(-1)`

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

Let `sin^(-1) x = alpha :. sin alpha = x`

`:. -2 sin^(-1) x = -2 alpha = cos^(-1) (1-x)`

or `cos 2 alpha =1 -x :. cos (- theta) = cos theta`

`:. 1-2 sin^2 alpha = 1-x`

Putting `sin alpha = x => 1-2x^2 =1-x`

or `2x^2 -x= 0 ` `x (2x -1) =0 :. x=0, 1/2`

But `x= 1/2` does not satisfy the equation

`:. x= 0`

`:.` Option (c) 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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