**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 – 4xยณ) ; x โ [1/2,-1/2]**

Ans. Considering Sin^{-1}x = ฮธ

sin ฮธ = x

Taking RHS ,we have

Sin^{-1}(3x – 4xยณ)

Putting the value of x = sin ฮธ

Sin^{-1}(3sin ฮธ – 4sin ยณฮธ)[since sin 3ฮธ =3sin ฮธ – 4sin ยณฮธ]

Sin^{-1} (sin 3ฮธ)

= 3ฮธ = 3Sin^{-1}x = LHS

Hence proved

Q2.**3cos ^{-1}x = cos^{-1}(3x – 4xยณ) ; x โ [1/2,1]**

Ans. Considering cos^{-1}x = ฮธ

cos ฮธ = x

Taking RHS ,we have

cos^{-1}(3x – 4xยณ)

Putting the value of x = cos ฮธ

cos^{-1}(3cosยณ ฮธ – 4cos ฮธ)[since cos 3ฮธ = 4cosยณฮธ -3cos ฮธ]

cos^{-1} (cos 3ฮธ)

= 3ฮธ = 3cos^{-1}x = LHS

Hence proved

**Q3. Prove thatย **

**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 ฮธ

We get

= ฮธ/2

**Q6. Write the function in the simplest form**

Ans. Considering x = cosec ฮธ

Tan^{-1}(tan ฮธ) = ฮธ

Since x = cosec ฮธโฮธ = cosec^{-1}x

**Q7. Write the function in the simplest form**

Ans. The given function is

Since 1 – cos x = 2sinยฒ(x/2) and 1 + cos x = 2cosยฒ(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(ฯ/4 – x)]

= ฯ/4 – x

**Q9. Write the function in the simplest form**

Ans. The given function is

Considering x = a sin ฮธ, we have

= tan^{-1 }(tan ฮธ) = ฮธ

Since, x = a sin ฮธ, hence sin ฮธ = x/a

ฮธ = sin^{-1}(x/a)

**Q10. Write the function in the simplest form**

Ans.The given function is

Considering x =ย a tan ฮธ, we have

= tan^{-1 }(tan 3ฮธ) = 3ฮธ

Since x = a tan ฮธ โ tan ฮธ = x/a , ฮธ = tan^{-1 }(x/a)

Therefore 3ฮธ = 3tan^{-1 }(x/a)

Find the value of each of the following:

Ans.

= tan^{-1}1 โ ฯ/4

cot ฯ/2 =0

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

**Q12. cot ( sin ^{-1}ฮฑ + cos^{-1}ย ฮฑ)**

Ans. We are given

cot ( sin^{-1}ฮฑ + cos^{-1}ย ฮฑ)

Since, sin^{-1}x + cos^{-1} x) = ฯ/2

cot ฯ/2 = 0

Ans. We are given the function

Considering x = tan A and y = tan B

= tan (A +ย B)

Since x = tan A and y = tan B

**Q14.If sin[ sin ^{-1}(1/5) + ย cos^{-1}x ] =1, then find the value of x**

Ans. We are given that

sin[ sin^{-1}(1/5) + ย cos^{-1}x ] =1

Substituting 1=Sin ฯ/2

sin[ sin^{-1}(1/5) + ย cos^{-1}x ] =sin ฯ/2

sin^{-1}(1/5) + ย cos^{-1}x = ฯ/2

Since sin^{-1}A + ย cos^{-1}A= ฯ/2

Therefore

cos^{-1}x = ฯ/2 – sin^{-1}(1/5)ย =cos^{-1}1/5

x = 1/5

Ans. We are given that

Applying the following formula

2xยฒ -4 = -3

2xยฒ = 1โ x = ยฑ1/โ2

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