Rajasthan Board RBSE Class 12 Maths Chapter 2 Inverse Circular Functions Miscellaneous Exercise
RBSE Solutions For Class 12 Maths Chapter 2 Miscellaneous Question 1.
Principal value of tan-1(- 1) is :
(a) 45°
(b) 135°
(c) – 45°
(d) – 60°
Solution:
∵ tan-1(- x)= – tan-1 x
tan-1 (-1) = – tan-1(1)
Let tan-11 = θ
∴ tan θ = 1
⇒ tan θ = tan 45°
∴ θ = 45°
∴ tan-1(-1) = – 45°
So, option (c) correct.
RBSE Class 12 Maths Chapter 2 Miscellaneous Question 2.
2 tan-1(1/2) is equal to :
Solution:
So, option (a) is correct.
RBSE Solutions For Class 12 Maths Chapter 2 Question 3.
If tan-1 (3/4) = θ, then sin is :
Solution:
According to question,
RBSE Solution Class 12 Maths Chapter 2 Question 4.
cot (tan-1 α + cot-1 α) is equal to:
(a) 1
(b) ∞
(c) 0
(d) none of those
Solution:
cos (tan-1 a + cot-1 a)
So, option (c) correct.
Class 12 Maths RBSE Solution Chapter 2 Question 5.
If sin-1 ( \(\frac { 1 }{ 2 }\) ) = x, then general value of x is :
Solution:
Given,
Hence, option (d) is correct.
RBSE Solutions For Class 12 Maths Chapter 2 Example Question 6.
2 tan (tan-1x + tan-1 x3) is :
Solution:
Hence, option (a) is correct.
Miscellaneous Exercise 2 Class 12 Question 7.
If tan-1(3x) + tan-1 (2x) = \(\frac { \pi }{ 4 }\), then x is:
Solution:
RBSE Class 12 Maths Chapter 2 Question 8.
Value of sin-1( \(\frac { \sqrt { 3 } }{ 2 }\) ) + 2 cos-1 (\(\frac { \sqrt { 3 } }{ 2 }\) )is :
Solution:
Inverse Circular Functions Class 12 RBSE Question 9.
If tan-1(1) + cos-1 ( \(\frac { 1 }{ \sqrt { 2 } } \)) = sin-1 x, then value of x is:
Solution:
Misc Ex Ch 2 Class 12 Question 10.
If cot-1 x + tan-1 \(\frac { 1 }{ 3 }\) = \(\frac { \pi }{ 2 }\) then x is :
(a) 1
(b) 3
(c) \(\frac { 1 }{ 3 }\)
(d) none of these
Solution:
Hence, option (c) is correct.
RBSE Class 12 Maths Chapter 2 Solution Question 11.
If 4 sin-1 x + cos-1x = π, then find x.
Solution:
4 sin-1x + cos-1x = π
Class 12 Maths Chapter 2 Miscellaneous Exercise Solutions Question 12.
Solution:
Class 12 Maths Ch 2 Miscellaneous Question 13.
If sin-1 (\(\frac { 3 }{ 4 }\) ) + sec-1 (\(\frac { 4 }{ 3 }\) ) = x, then find x.
Solution:
Chapter 2 Maths Class 12 Question 14.
Find: sin-1 (\(\frac { 4 }{ 5 }\)) + 2 tan-1 (\(\frac { 1 }{ 3 }\))
Solution:
RBSE Solutions For Class 12 Maths Chapter 11 Miscellaneous Question 15.
If sin-1(\(\frac { 5 }{ 13 }\)) + sin-1 (\(\frac { 12 }{ x }\)) = 90°, then find x.
Solution:
Miscellaneous Exercise On Chapter 2 Class 12 Question 16.
Prove that:
Solution:
Ex 2 Miscellaneous Class 12 Question 17.
tan-1x + tan-1y + tan-1z = π, then prove x + y + z = xyz
Solution:
tan-1x + tan-1y + tan-1z = π
Chapter 2 Class 12 Miscellaneous Question 18.
Prove that tan-1(\(\frac { 1 }{ 2 }\) tan 2A) + tan-1 (cot A) + tan-1 (cot2 A) = 0.
Solution:
Miscellaneous Chapter 2 Class 12 Question 19.
Prove that tan-1x = 2 tan-1(cosec (tan-1 x) – tan (cot-1 x)].
Solution:
Let tan-1x = θ
Ch 2 Maths Class 12 Miscellaneous Question 20.
then prove that value of Φ – θ is 30°.
Solution:
Question 21.
Prove that:
Solution: