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NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

ex.2.2 inverse trigonometric functions

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.

ex.2.2 inverse trigonometric functions

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

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Q1.3Sin-1x = Sin-1(3x – 4x³) ; x ∈ [1/2,-1/2]

Ans. Considering Sin-1x = 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-1x = LHS

Hence proved

Q2.3cos-1x = cos-1(3x – 4x³) ; x ∈ [1/2,1]

Ans. Considering cos-1x = 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-1x = LHS

Hence proved

Q3. Prove that 

tan-12/11 + tan-17/24= tan-11/2

Taking the LHS

tan-12/11 + tan-17/24

Applying the formula

= Tan-1(125/250)

= Tan-11/2 = 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-1x

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-11 ⇒ π/4

cot π/2 =0

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

Q12. cot ( sin-1α + cos-1 α)

Ans. We are given

cot ( sin-1α + cos-1 α)

Since, sin-1x + 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-1x ] =1, then find the value of x

Ans. We are given that

sin[ sin-1(1/5) +  cos-1x ] =1

Substituting 1=Sin π/2

sin[ sin-1(1/5) +  cos-1x ] =sin π/2

sin-1(1/5) +  cos-1x = π/2

Since sin-1A +  cos-1A= π/2

Therefore

cos-1x = π/2 – sin-1(1/5)  =cos-11/5

x = 1/5

Ans. We are given that

Applying the following formula

2x² -4 = -3

2x² = 1⇒ x = ±1/√2

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