Mathematics Miscellaneous Examples and Miscellaneous Exercise

### Miscellaneous Examples

Q 3143778643

Find the value of sin ^(-1) (sin \ (3 pi)/5)

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
Q 3103778648

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

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
Q 3163878745

Show that sin^(-1) \ 12/13 + cos^(-1) \ 4/5 + tan^(-1) \ 63/16 = pi

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
Q 3103878748

Simplify tan^(–1) [(a cos x- b sin x)/(b cos x + a sin x)], if a/b tan x > -1

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
Q 3143078843

Solve tan^(–1) 2x + tan^(–1) 3x = pi/4

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.

### Miscellaneous Exercise

Q 2615178960

Find the value of the following:

cos^(-1) (cos (13 pi )/6)
Class 12 Exercise 2.mis Q.No. 1
Solution:

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

= cos^(-1) cos pi/6 = pi/6
Q 2625178961

Find the value of the following:

tan^(-1) (tan ((7 pi)/6))
Class 12 Exercise 2.mis Q.No. 2
Solution:

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

= tan^(-1) tan (pi/6) = pi/6
Q 2655178964

Prove that sin^(-1 ) 8/17 + sin^(-1) 3/5 = tan^(-1) 77/36

Class 12 Exercise 2.mis Q.No. 4
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
Q 2665178965

Prove that cos^(-1) (4/5) +cos^(-1) (12/13) = cos^(-1) (33/65)
Class 12 Exercise 2.mis Q.No. 5
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)
Q 2685178967

Prove that cos^(-1) (12/13) + sin^(-1) (3/5) = sin^(-1) (56/65) 
Class 12 Exercise 2.mis Q.No. 6
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)
Q 2605178968

Prove that cos^(-1) (12/13) + sin^(-1) (3/5) = sin^(-1) (56/65) 
Class 12 Exercise 2.mis Q.No. 6
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)
Q 2615178969

Prove that tan^(-1) (63/16) = sin^(-1) (5/13) + cos^(-1) (3/5)
Class 12 Exercise 2.mis Q.No. 7
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)
Q 2615180060

Prove that:

tan^(-1) (1/5) + tan^(-1) (1/7) + tan^(-1) (1/3) + tan^(-1) (1/8) = pi/4
Class 12 Exercise 2.mis Q.No. 8
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
Q 2645180063

Prove that tan^(-1) sqrt (x) = 1/2 cos^(-1) ( (1-x)/(1+x) ) x in [0,1]
Class 12 Exercise 2.mis Q.No. 9
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.
Q 2655180064

cot^(-1) ( ( sqrt (1+ sin x) + sqrt (1- sin x ) )/( sqrt (1+sin x) - sqrt (1- sin x) ) ) = x/2 , x in (0, pi/4)
Class 12 Exercise 2.mis Q.No. 10
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
Q 2665180065

tan^(-1) ( ( sqrt (1+x) - sqrt (1-x) )/( sqrt (1+x) + sqrt (1-x) ) ) = pi/4 -1/2 cos^(-1) x,

-1/2 cos^(-1) x, - 1/sqrt(2) le x le 1

(Hint: Put x = cos 2 theta ]
Class 12 Exercise 2.mis Q.No. 11
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
Q 2675180066

(9 pi)/8 - 9/4 sin^(-1) (1/3) = 9/4 sin^(-1) ( (2 sqrt (2) )/3) 
Class 12 Exercise 2.mis Q.No. 12
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 )
Q 2685180067

Solve the following equations:

2 tan^(-1) (cos x) = tan^(-1) (2 cosec x)
Class 12 Exercise 2.mis Q.No. 13
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
Q 2605180068

tan^(-1) \ \ (1-x)/(1+x) =1/2 tan^(-1) x, (x > 0 )
Class 12 Exercise 2.mis Q.No. 14
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)
Q 2615180069

sin (tan^(-1) x), |x| < 1 is equal to

Class 12 Exercise 2.mis Q.No. 15
(A)

x/(sqrt (1-x^2) )

(B)

1/(sqrt (1-x^2) )

(C)

1/(sqrt (1+x^2) )

(D)

x/(sqrt (1+x^2) )

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.
Correct Answer is => (D) x/(sqrt (1+x^2) )
Q 2615280160

sin^(-1) (1 - x)- 2 sin^(-1 ) x = pi/2, then x is equal to
Class 12 Exercise 2.mis Q.No. 16
(A)

0, 1/2

(B)

1,1/2

(C)

0

(D)

1/2

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.
Correct Answer is => (C) 0
Q 2625280161

tan^(-1) (x/y) - tan^(-1) ( (x-y)/(x+y)) is equal to
Class 12 Exercise 2.mis Q.No. 17
(A)

pi/2

(B)

pi/3

(C)

pi/4

(D)

(-3 pi )/4

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.
Correct Answer is => (C) pi/4`