NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions
NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions will give you an idea of the technics used in solving the different types of questions based on the inverse trigonometric functions. The NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions are solved and scrutinized by a team of experts. The purpose of these NCERT solutions is to help students in clearing their doubts so that they could achieve excellent marks in the exam.
NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions
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Q1.3Sin^{-1}x = Sin^{-1}(3x – 4x³) ; x ∈ [1/2,-1/2]
Ans. Considering Sin^{-1}x = θ
sin θ = x
Taking RHS ,we have
Sin^{-1}(3x – 4x³)
Putting the value of x = sin θ
Sin^{-1}(3sin θ – 4sin ³θ)[since sin 3θ =3sin θ – 4sin ³θ]
Sin^{-1} (sin 3θ)
= 3θ = 3Sin^{-1}x = LHS
Hence proved
Q2.3cos^{-1}x = cos^{-1}(3x – 4x³) ; x ∈ [1/2,1]
Ans. Considering cos^{-1}x = θ
cos θ = x
Taking RHS ,we have
cos^{-1}(3x – 4x³)
Putting the value of x = cos θ
cos^{-1}(3cos³ θ – 4cos θ)[since cos 3θ = 4cos³θ -3cos θ]
cos^{-1} (cos 3θ)
= 3θ = 3cos^{-1}x = LHS
Hence proved
Q3. Prove that
tan^{-1}2/11 + tan^{-1}7/24= tan^{-1}1/2
Taking the LHS
tan^{-1}2/11 + tan^{-1}7/24
Applying the formula
= Tan^{-1}(125/250)
= Tan^{-1}1/2 = RHS, Hence proved
NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions
Q4. Prove that
2tan^{-1}(1/2) +tan^{-1}(1/7) =tan^{-1}(31/17)
Taking the LHS
2tan^{-1}(1/2) +tan^{-1}(1/7)
Reawritting it as following
tan^{-1}(1/2)+tan^{-1}(1/2) +tan^{-1}(1/7)
Applying the formula
= tan^{-1}(1/2) +tan^{-1}(9/13)
Again applying the same formula
= tan^{-1}(31/17) = RHS, Hence proved
Q5.Write the function in the simplest form
Ans. Considering the value of x = tan θ
We get
= θ/2
Q6. Write the function in the simplest form
Ans. Considering x = cosec θ
Tan^{-1}(tan θ) = θ
Since x = cosec θ⇒θ = cosec^{-1}x
Q7. Write the function in the simplest form
Ans. The given function is
Since 1 – cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2)
=tan^{-1}(tan x/2) = x/2
NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions
Q8. Write the function in the simplest form
Ans. The given function is
For writting the given function in the form of tan,dividing the numerator and denominator by cos x
Writing it in the following form
= tan^{-1 }[tan(π/4 – x)]
= π/4 – x
Q9. Write the function in the simplest form
Ans. The given function is
Considering x = a sin θ, we have
= tan^{-1 }(tan θ) = θ
Since, x = a sin θ, hence sin θ = x/a
θ = sin^{-1}(x/a)
Q10. Write the function in the simplest form
Ans.The given function is
Considering x = a tan θ, we have
= tan^{-1 }(tan 3θ) = 3θ
Since x = a tan θ ⇒ tan θ = x/a , θ = tan^{-1 }(x/a)
Therefore 3θ = 3tan^{-1 }(x/a)
Find the value of each of the following:
Ans.
= tan^{-1}1 ⇒ π/4
cot π/2 =0
NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions
Q12. cot ( sin^{-1}α + cos^{-1} α)
Ans. We are given
cot ( sin^{-1}α + cos^{-1} α)
Since, sin^{-1}x + cos^{-1} x) = π/2
cot π/2 = 0
Ans. We are given the function
Considering x = tan A and y = tan B
= tan (A + B)
Since x = tan A and y = tan B
Q14.If sin[ sin^{-1}(1/5) + cos^{-1}x ] =1, then find the value of x
Ans. We are given that
sin[ sin^{-1}(1/5) + cos^{-1}x ] =1
Substituting 1=Sin π/2
sin[ sin^{-1}(1/5) + cos^{-1}x ] =sin π/2
sin^{-1}(1/5) + cos^{-1}x = π/2
Since sin^{-1}A + cos^{-1}A= π/2
Therefore
cos^{-1}x = π/2 – sin^{-1}(1/5) =cos^{-1}1/5
x = 1/5
Ans. We are given that
Applying the following formula
2x² -4 = -3
2x² = 1⇒ x = ±1/√2
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