Revision/test (Inverse Trigonometry)

TEST

1) Prove: sin⁻¹(8/17) + sin⁻¹(3/5) = sin⁻¹(77/85).                              (2)

2) Evaluate: sin(3 sin⁻¹0.4).        (2)

3) If x, y, z belongs to [-1, 1] such that sin⁻¹x + sin⁻¹y + sin⁻¹z = 3π/2, find the value of x²⁰²⁰ + y²⁰²¹+ z²⁰²² - 9/(x²⁰²⁰+ y²⁰²¹ + z²⁰²²).              (3)

4) Prove: sin[cot⁻¹{cos(tan⁻¹x)}] = √{(x²+1)/(x²+2)}.                       (5) 

OR

Tan⁻¹(1/2 tan 2A) + tan⁻¹(cotA) + tan⁻¹(cot³A). Evaluate 

5) Compulsory

A) The value of sin(1/4 sin⁻¹(√63/8) is

a) 1/√2 b) 1/√3 c) 1/2√2 d) 1/3√3

B) If tan⁻¹(cot A)= 2A, then A=

a) ±π/3 b) ±π/4 c) ±π/6 d) none

C) 4 cos⁻¹x + sin⁻¹x =π, then the value of x is

b) 3/2 b)1/√2 c) √3/2 d) 2/√3

REVISION 

Prove

1) 2tan⁻¹1/5 + tan⁻¹1/8=tan⁻¹4/7

2) 1/2 tan⁻¹x = 

  cos⁻¹√[{1√(1+x²)}/2√(1+x²)]

3) sin⁻¹4/5 +sin⁻¹16/65= π/2

4) sin (π/3 - sin⁻¹(-1/2)) = 1

5) cot (π/4 - 2 cot⁻¹3) =7

6) If cos⁻¹x+ cos⁻¹y+ cos⁻¹z=π then prove x²+y²+z²+2xyz=1

7) If tan⁻¹x+tan⁻¹y+tan⁻¹z=π/2 then prove xy +yz+zx =1

8) If tan⁻¹x+ tan⁻¹y+ tan⁻¹z = 0 then prove x+y+z=xyz

9) prove tan⁻¹(1/2 tan  2A)  + tan⁻¹(cot A)+ tan⁻¹(cot³A) =0

10)a) solve: tan⁻¹3x+ tan⁻¹2x =π/4

b) tan⁻¹(cos x) =tan⁻¹(2 cosec x)

c) tan⁻¹(x+1) +tan⁻¹(x-1) =tan⁻¹8/31

d) sin⁻¹x/√(1+x²) - sin⁻¹1/√(1+x²) = sin⁻¹{(1+x)/(1+x²)}

e) Sin⁻¹(x/2)+cos⁻¹(x+√3/2) =π/6

f) If Sin⁻¹x +Sin⁻¹y+Sin⁻¹z=3π/2, then prove x²+y²+z²+2xyz=1

Simplify:

11) a) Cos⁻¹(x+ 1/2) +Cos⁻¹x+ Cos⁻¹(x - 1/2) = 3π/2.

b) Tan⁻¹{(2sin2θ)/1+2cos2θ)} - 1/2  sin⁻¹{(3sin2θ)/(5+4cos2θ)}

c) tan⁻¹1 + tan⁻¹1/2 + tan⁻¹1/3

d) sin(sin⁻¹1/3 + sec⁻¹3) + cos(tan⁻¹1/2 + tan⁻¹2)

e) sin{sin⁻¹√5/4 + tan⁻¹√(5/11)}

f) tan⁻¹sin cos⁻¹√(2/3)

Prove:

12)a) cos⁻¹3/5+ cos⁻¹12/13 + cos⁻¹63/65  = π/2

b) 1/2 tan⁻¹x = cos⁻¹√{(1+√(1+x²))}/2√(1+x²)

c) prove tan⁻¹x + cot⁻¹(x+1) = tan⁻¹(x²+x+1)

d) tan(2 tan⁻¹m) = 2tan(tan⁻¹m+ tan⁻¹m³)

e) tan⁻¹(1/2  tan 2A) + tan⁻¹(cotA)+ tan⁻¹(cot³A) = 0

Solve:

13)a) 3(Cos⁻¹x + 2sin⁻¹x) = 7π

b) tan⁻¹(cot2x) + tan⁻¹(-cot 3x) = x

c) sin⁻¹{2a/(1+a²)} - cos⁻¹{(1-b²)/(1+b²) = 2 tan⁻¹x

d) tan⁻¹x tan⁻¹2x tan⁻¹3x = π

e) 3 tan⁻¹{1/(2+√3)} - tan⁻¹( = 1/x) = tan⁻¹1/3

f) sin⁻¹x + sin⁻¹(1-x) = cos⁻¹x

g) tan⁻¹(x+1)+tan⁻¹(x-1)= tan⁻¹8/31

h) tan⁻¹{(x-1)/(x-2)} + tan⁻¹{(x+1)/(x+2)} = π/4

i) tan⁻¹{(1/(2x+1)}+ tan⁻¹{1/(4x+1)= tan⁻¹(2/x²)

14)  if α+β= tan⁻¹m, α-β=tan⁻¹n Express tan 2α and tan2β in terms of m,n.

15) If  sin⁻¹x = tan⁻¹y then find the value of 1/x² - 1/y²

16) If r²= x²+y²+z², prove tan⁻¹yz/xr + tan⁻¹zx/yr + tan⁻¹xy/zr = π/2

Solve:

17)a) cos(2tan⁻¹1/7) =  sin(4tan⁻¹x)

b) cos(tan⁻¹x) = sin(cot⁻¹3/4).

prove :

18)a) cos⁻¹(63/65)+2tan⁻¹(1/5) = sin⁻¹(3/5).

  

b) 4 tan⁻¹1/5 - tan⁻¹1/70 + tan⁻¹1/99  =π/4

c) sin⁻¹{x/√(1+x²)} +   cos⁻¹{(x+1) √(x²+2x+2)} =  tan⁻¹(x²+x+1)

d) cot(π/4 - 2 cot⁻¹3) = 7

e) sin cot⁻¹ cos(tan⁻¹x) = √{(1+x²)/(2+x²)

f) sin⁻¹√3/2 + 2 tan⁻¹1/√3 = 2π/3

g) tan⁻¹1/3 +tan⁻¹1/5 + tan⁻¹1/7 + tan⁻¹1/8 = π/4

h) tan⁻¹(1/2 tan 2A) +tan⁻¹(cot A)+ tan⁻¹(cot³A) = 0

Solve:

19)a) tan⁻¹(2+x)+ tan⁻¹(2-x) =tan⁻¹2/3

b) sin⁻¹5/x + sin⁻¹12/x = π/2

c) sin⁻¹6x + sin⁻¹(6√3 x)= - π/2

d) sin⁻¹{2a/(1+a²)} + sin⁻¹{2b/(1+b²)} = 2 tan⁻¹x

e) sin{2 cos⁻¹ cot(2 tan⁻¹x)}=0

20) If tan⁻¹a + tan⁻¹b + tan⁻¹c = π then Prove that a+ b + c = abc

21) If tan⁻¹a + tan⁻¹b + tan⁻¹c = π/2 prove ab + bc + ca = 2

22) If cos⁻¹a/2 + cos⁻¹b/3  = K prove 9x² - 12 xy cos K + 4y² = 36 sin²K

Prove:

23)a) tan⁻¹(1/2) + tan⁻¹(1/3) = π/4

b) 4tan⁻¹1/5 + tan⁻¹1/70 + tan⁻¹1/89 =π/4

c) tan⁻¹(1/4)+tan⁻¹2/9= 1/2cos⁻¹3/5

d) sin⁻¹4/5 + sin⁻¹5/13 + sin⁻¹16/65 =π/2

e) 1/2 tan⁻¹x = cos⁻¹√[{(1+√(1+x²)}/2√(1+x²)]

f) tan⁻¹x + cot⁻¹(x+1)= tan⁻¹(x²+x+1)

g) tan⁻¹1+tan⁻¹2+tan⁻¹3 = 0

h) 2(tan⁻¹1+tan⁻¹1/2 +tan⁻¹1/3) =π

i) tan(2tan⁻¹a) = 2tan(tan⁻¹a+ tan⁻¹a³)

j)  tan⁻¹(1/2 tan 2A)+ tan⁻¹(cot A)+ tan⁻¹(cot³A) =0

Solve:

24)a) tan⁻¹(x+1)+tan⁻¹(x-1) = tan⁻¹8/31

b) tan⁻¹(x-1)+tan⁻¹x+tan⁻¹(x+1) = tan⁻¹3x

c) tan⁻¹{(x-1)/(x-2)}+ tan⁻¹{(x+1)/(x+2)} = π/4

25) If cos⁻¹x + cos⁻¹y + cos⁻¹z =π prove x²+y²+z²+2xyz =1

26) prove sin⁻¹(12/13) + cos⁻¹(4/5)  + tan⁻¹(63/16) = π .

Solve :

27)a) cos⁻¹(sin cos⁻¹x) = π/6

b) solve) cos (sin⁻¹x)= 1/7

c) value of tan(2 tan⁻¹ 1/5)

d) tan⁻¹{1/(2x+1)} +  tan⁻¹{1/(4x+1)} = tan⁻¹2/x²

e) tan⁻¹{(x+1)/(x-2)} +tan⁻¹{(x-1)/x} = tan⁻¹(-7)

28) If tan⁻¹x +tan⁻¹y+tan⁻¹z= π prove x+y+z=xyz

29) if tan⁻¹x+tan⁻¹y+tan⁻¹z=π/2 show that xy+yz+zx=1

29)  prove tan(π/4 + 1/2cos-¹ 1/3) + Tan(π/4 - 1/2 cos-¹ 1/3) =6

 30) solve cos⁻¹(sin (cos⁻¹x))=π/3. 

                                         

 Prove :

31) a) Sin(2tan⁻¹3/5-sin⁻¹7/25) =304/425

32) If tan⁻¹x + tan⁻¹y + tan⁻¹z=0, prove that x+ y +z = xyz.

33) prove: tan⁻¹1/4 + tan⁻¹2/9 = 1/2 cos⁻¹3/5

Revision/Test(Compound Angle)

1) Prove: tan 11A tan 7A tan 4A = tan 11A - tan 7A - tan 4A.

2) If sin(x+ y)= 4/5, cos(x - y)= 12/13, Find tan 2x.

3) Show: cos 2x cos 2y + sin²(x - y)- sin²(x + y)= cos(2x + 2y).

4) Prove: 4 sin(π/3  - x) sin(π/3 + x)= 3 - 4 sin²x.

5) Show that tan 75= {(√3+ 1)/(√3 - 1)}= 2+ √3. Hence show that tan 75 - cot 75 = 4 sin60.

Revision/Test (SET THEORY)

1) If A={1,2,3,4,5,6,7,9,11,13,15} and B={2,4,6,8,10,12,14,16} find A - B and B - A. 

2) The production manage of a company examined 100 items and furnished the following report:

Defect in measuring 50, defect in coloring 30, defect in quality 23, defect in measurement and coloring 8, defect in quality and measurement 20, defect in quality and coloring 10 and 5 are defective in all respects. The manager was penalised for the report. Use set theory to explain the reason for penal measure.

3) If A={1,2,3,4} and B={3,4,5} and C={1,4,5} then verify find A - (B U C)= (A- B)∩(A - C).

4) A market research group conducted a survey of 1000 customers and reported that 700 consumers like product A and 480 consumers like product B. What is the least number that must have liked both the products?

5) If A={1,2,3,4} and B={2,4,5, 6} and C={1,3,4,6,8} then find A ∩(B U C).

6) In a class of 75 students, 48 read Bengali, 33 read mathematics and 19 read both the subjects. How many of them read none of these subjects!

7) If A={1,2,3,4} and B={2,4,5, 6} and C={3,4,6} then find A - (B U C) and (A- (B∩ C).

8) If A={x, y, z{ then write down the power set of A.

9) If S={1,2,3,4,5,6,7,8,9,10}, A={1,2,3,4} and B={2,4,6,8,10} then find (A UB)' And (A∩B)'

10) Three sets A, B, C be such that A- B={2,4,6}; A- C={2,3,5}; find 

  A - (B U C) and A- (B∩C).

11) Using set theory, find the LCM and HCF of 4, 6 and 8.

12) If A={p,q,r,s} and B={r,s,t} and C={p,s,t} then verify find A - (B U C)= (A- B)∩(A - C).

13) If A={1,4} and B={4,5} and C={5,7} then find (A x B) U(A x C)

14) If A={x:x is an integer and 1 ≤ x ≤ 10} and B={x:x is a multiple of 3 and 5 ≤ x ≤ 30}, find A∩B.

15) Using set theory find the HCF of 15, 40, 105.

16) Find the power set of {2,3,5}.

17) If A={1,2,3} and B={4,6,8} then find A x B.

18) If A={x:x is a natural number and x ≤ 6} and B={x:x is the natural numbers and 3 ≤ x ≤ 8}, find A - B and A∩B.

19) In a class of 100 students, 55 students studied history, 41 students studied English and 25 students studied both the subjects. Find the number of students who study neither of the subjects.

20) If A={1,2} find the power set of A.

21) If A={1,2,3,4} and B={2,4,5,8} and C={3,45,6,7} then find A U (B ∩C) and A ∩(B U C).

22) Each students in a class of 40, plays atleast one of the games hockey, football and cricket; 16 play hockey, 20 football, 26 cricket, 5 play hockey and football, 14 football and cricket and hockey, football and cricket. Find the number of students who play hockey and cricket but not football.

23) If A={x: x²- 5x +6= 0}, B={2,4} and C={4,5} then AX (B ∩ C) is

A) a null set 

Revision/Test(Quadratic equation)

1) 36x²+ 23= 60x. 

2) 3x² -17 x +25= 0. 

3) 15x² -28 x = x.

4) x² + 3x - 3= 0 (upto 2 decimal places). 

5) (x² + 8)/11= 5x - x² - 5. 

6) 2x/(x- 4) + (2x - 5)/(x - 3)= 25/3. 

7) (x+ 6)/(x- 4)- (x + 1)/(x +2)= 1/(3x +1). 

8) (x+1)(x+2)(x+3)(x+4) + 1= 0. 

9) x(x- 1)(x+2)(x- 3)(x+4) + 8= 0.    

10) √(x² - 3x) = 4x² - 12x -3.     

11) √{(x²+2)/x²-2)}+ 6√{(x²-2)/(x²+2)}= 5.

12) Find the nature of following:

a) x² - 5x -2= 0.

b) 4x² - 4x +1= 0.

c) x² + px - q²= 0.

13) Find the value of m so that the roots of the equation (4- m)x² +(2m +4)x + (8m +1)= 0 may be equal. 0,3

14) If (1- p) is a root of the equation x² + px + (1- p) = 0. Then its roots are

a) 0, -1 B) -1,1 C) 0,1 D) -1,2

15) √{x/(1- x)} + {(1- x)/x} = 13/6.

16) (x+1)(x+2)(x+3)(x+4) = 120.  

17) 2x² + 2x -3= 0 (upto two decimal places).