Inverse Trigonometric Functions Practice Questions - DPP for JEE

 Page 2


 if 
Also, 
 
Hence, 
,
= 
[Since, ]
3. (b) We have, 
? 
? 
? 
? 
4. (d) We have
Page 3


 if 
Also, 
 
Hence, 
,
= 
[Since, ]
3. (b) We have, 
? 
? 
? 
? 
4. (d) We have
Put x = tan ? and y = tan f, we get
? cos
–1
 (cos 2?) + cos
–1
 (cos 2f) = 
? 
So, tan
–1
 x + tan
–1
 y = 
? x + y = 1 – xy ? x + y + xy = 1
5. (c) sin
–1
 (log[x]) is defined if  and 
   [x] = 1, 2  
Again, is defined if
 and 
 Domain of 
For  [x] = 1
  
 Range of 
6. (b) We have, 
Page 4


 if 
Also, 
 
Hence, 
,
= 
[Since, ]
3. (b) We have, 
? 
? 
? 
? 
4. (d) We have
Put x = tan ? and y = tan f, we get
? cos
–1
 (cos 2?) + cos
–1
 (cos 2f) = 
? 
So, tan
–1
 x + tan
–1
 y = 
? x + y = 1 – xy ? x + y + xy = 1
5. (c) sin
–1
 (log[x]) is defined if  and 
   [x] = 1, 2  
Again, is defined if
 and 
 Domain of 
For  [x] = 1
  
 Range of 
6. (b) We have, 
7. (c) Let S
8
 = cot
–1
2 + cot
–1
 8 + cot
–1
 18 + cot
–1
 32 + ....
? T
n
 = cot
–1
 2n
2
= tan
–1
 (2n + 1) – tan
–1
 (2n – 1)
? = tan
–1
 8 – tan
–1
 1
8. (d) Let cos
–1
 x + cos
–1
 y = 
Page 5


 if 
Also, 
 
Hence, 
,
= 
[Since, ]
3. (b) We have, 
? 
? 
? 
? 
4. (d) We have
Put x = tan ? and y = tan f, we get
? cos
–1
 (cos 2?) + cos
–1
 (cos 2f) = 
? 
So, tan
–1
 x + tan
–1
 y = 
? x + y = 1 – xy ? x + y + xy = 1
5. (c) sin
–1
 (log[x]) is defined if  and 
   [x] = 1, 2  
Again, is defined if
 and 
 Domain of 
For  [x] = 1
  
 Range of 
6. (b) We have, 
7. (c) Let S
8
 = cot
–1
2 + cot
–1
 8 + cot
–1
 18 + cot
–1
 32 + ....
? T
n
 = cot
–1
 2n
2
= tan
–1
 (2n + 1) – tan
–1
 (2n – 1)
? = tan
–1
 8 – tan
–1
 1
8. (d) Let cos
–1
 x + cos
–1
 y = 
?  = 
? sin
–1
 x + sin
–1
 y = .
9. (a) Since, 
10. (d) Let sin
–1
 a = x ? a = sin x
sin
–1 
b = y ? b = sin y
sin
–1 
c = z ? c = sin z
?
= sin x cos x + sin y cosy + sinz cosz
= (1/2) (sin2x + sin2y + sin 2z) =(1/2) (4sin x sin y sin z)
= 2 sinx siny sinz = = 2abc
11. (a) We have
Since  for all x