Chapter-2 (Inverse trigonometric function) Session 2016-17

CHAPTER 2
DAV CENTENARY PUBLIC SCHOOL, PASCHIM ENCLAVE, NEW DELHI-87
CLASS-XII
SESSION 2016-17
Inverse trigonometric function.Expected marks : 5 (1 + 4)
Expected No. of Questions : 2
One Mark Questions
Q1. Write the principle value of following.
7πœ‹
3πœ‹
βˆ’1
a) cos-1cos 6
b) sin-1 sin 5
c) cos -1 ( 2 )
3πœ‹
βˆ’πœ‹
e)sin-1tan 4
f) cos-1cos 4
d) tan-1(-1)
βˆ’πœ‹
g) sec2(tan-12)
h) cot-1 cot
4
1
Q2. If sin { sin-15 + cos-1 x } = 1 . Find x
Q3. Evaluate : a) sin[
πœ‹
3
βˆ’1
2πœ‹
- sin-1( 2 ) ]
2πœ‹
b) cos-1cos ( 3 ) + sin-1sin( 3 )
Q4. Write one branch of tan-1x other than principle branch.
Q5. Simplify the following expressions:
a) tan-1
1βˆ’π‘π‘œπ‘ π‘₯
π‘Žπ‘ +1
1+π‘π‘œπ‘ π‘₯
π‘₯
π‘₯βˆ’π‘¦
π‘₯βˆ’1
c) tan-1𝑦 - tan-1π‘₯+𝑦
e) tan-1
𝑏𝑐 +1
π‘Žπ‘ +1
b) cot-1 π‘Ž βˆ’π‘ + cot-1 π‘βˆ’π‘ + cot-1 π‘Žβˆ’π‘
π‘₯+1
d) tan-1π‘₯+1
π‘₯
+ tan-1π‘₯βˆ’1
f) cos 2tan-1
π‘Ž 2 βˆ’π‘₯ 2
1βˆ’π‘₯
1+π‘₯
Four marks questions
Q1. Prove the following:
βˆ’4
12
a) cos tan-1 3 + sin-1 13
=
63
1
1
1
b) 4 tan-1 5 - tan-1 70 + tan-199 =
65
πœ‹
c) tan-1 1 + tan-1 2 + tan-1 3 = 2( cot-1 1 + cot-1 2 + cot-1 3)
4
5
16
d) sin-1 5 + sin-1 13 + sin-165 =
4
1
πœ‹
f) sin-1 5 + 2tan-1 3 = 2
πœ‹
2
1
2
12
4
1
3
e) tan-1 4 + tan-1 9 = 2cos-1 5
63
g) sin-1 13 + cos-1 5 +tan-116 = πœ‹
4
Q2. Draw the graph of a) f(x)= tan-1 x
b) g(x) = cosec-1x.
Q3. Simplify the following inverse trigonometric expressions:
π‘π‘œπ‘ π‘₯
a) cot -1( 1 + π‘₯ 2 + π‘₯)
πœ‹
1
π‘Ž
c) tan ( 4 + 2cos-1 𝑏 )
3π‘₯βˆ’4 1βˆ’π‘₯ 2
d) sin-1
5
f) cosec-1
π‘Ž 2 +π‘₯ 2
π‘₯
b) tan-11+𝑠𝑖𝑛π‘₯
πœ‹
1
π‘Ž
+ tan ( 4 - 2cos-1 𝑏 )
π‘₯
e) sin-1
9+π‘₯ 2
g) tan-1
.
1+π‘₯ 2 + 1βˆ’π‘₯ 2
1+π‘₯ 2 βˆ’ 1βˆ’π‘₯ 2
Q4. Solve for x :
1
π‘₯βˆ’1
a) tan(cos-1x) = sin(cot -1 2)
π‘₯ 2 βˆ’1
2π‘₯
c) cos-1π‘₯ 2 +1 + tan-1π‘₯ 2 βˆ’1 =
2πœ‹
1+π‘₯
d) tan-11βˆ’π‘₯ =
3
πœ‹
1+π‘₯ 2 + 1
tan-1x = cos-1
2
2 1+π‘₯ 2
4π‘₯(1βˆ’π‘₯ 2 )
Q6. Show that 4tan-1x = tan-11βˆ’6π‘₯ 2 +π‘₯ 4
π‘₯
𝑦
π‘₯βˆ’π‘¦
π‘₯+𝑦
πœ‹
4
+ tan-1x
f) tan-1(x+1) + tan-1(x-1) = tan-1 31
1
Q7. Show that 2tan-1
πœ‹
8
e) sin-1(1-x) – 2 sin-1x = 2
Q5. Show that
π‘₯+1
b) tan -1π‘₯βˆ’2 + tan-1π‘₯+2 = 4
πœƒ
𝑦 +π‘₯π‘π‘œπ‘ πœƒ
tan 2 = cos-1π‘₯+π‘¦π‘π‘œπ‘ πœƒ
Q8. If cos-1π‘Ž + cos-1 𝑏 = z. Show that
π‘₯2
π‘Ž2
-
2π‘₯𝑦
π‘Žπ‘
𝑦2
cosz + 𝑏 2 = sin2z .
π‘₯
Q9. If y = cot-1 π‘π‘œπ‘ π‘₯ - tan-1 π‘π‘œπ‘ π‘₯. Prove that sin y = tan2 2