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NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions

class 12 maths ex.2.1

NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions

NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions gives you an idea about a new trigonometric function which is known as an inverse trigonometric function, trigonometric functions like sin,cos, tan etc are defined as the ratio between the sides of a right triangle,if we are given the sides of right triangle ΔABC then we can easily get the value of tan A, sin B, etc.If we are given the ratio between the sides then how can we calculate the ∠A and ∠B if it is given that ∠C is 90°. For solving this problem the concept of opposite trigonometric function is raised known as Inverse Trigonometry Functions,if sin A = 3/4 then A =sin-1(3/4),we can see the value of A in the trigonometric table or evaluate with the help of the calculator.

All NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions are solved by an expert of maths by a step by step method,so every student can understand the solutions easily.

class 12 maths ex.2.1

NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions

Find the principal values of the following.

Ans.considering the given function equivalent to y

sin y = -1/2

sin π/6 = 1/2

sin y= -sin π/6

sin y = sin(-π/6)

y = -π/6

We know the range of Sin-1  is [-π/2,π/2]

So, the principal value of y is ( -π/6)

Let y = cos-1(√3/2)

cos y = √3/2

cos y = cos π/6

y  = π/6

The range of principal value of  cos-1  is [0, π]

So, the required principal value is π/6

Let y = cosec-1(2)

cosec y = 2

since cosec π/6 = 2

cosec y = cosec π/6

We know the range of Sin-1  is [-π/2,π/2]

Therefore the principal value of cosec-1(2) is π/6

Let y = tan-1(√3)

tan y = √3

tan y = tan π/6

The range of principal value of tan-1  is [-π/2,π/2]

Therefore the principal value of tan-1  √3 is π/6

Let y = cos-1  (-1/2)

cos y = -1/2

since cos π/3 = 1/2

cos y = – cos π/3 = cos (π-π/3)

cos y = cos (2π/3)

We, know the range of principal value of cos-1  (-1/2) is [0,π]

Hence, the principal value of  cos-1  (-1/2) is 2π/3

Let y = tan-1  (-1)

tan y = -1

since tan π/4 = 1

tan y = – tan π/4

tan y = tan (- π/4)

We, know the range of principal value of tan-1   is [-π/2,π/2]

Hence, the principal value of  tan-1  (-1) is -π/4

Let y = sec-1  (2/√3)

sec y = 2/√3

since sec π/6 = 2/√3

sec y =  sec π/6

We, know the range of principal value of sec-1   is [0,π]

Hence, the principal value of  sec-1  (-1/2) is π/6

Let y = cot-1  (√3)

cot y = √3

since cot π/6 = √3

cot y =  cot π/6

We, know the range of principal value of cot-1   is [-π/2,π/2]

Hence, the principal value of  cot-1  (√3) is π/6

Let y = cos-1  (-1/√2)

cos y = -1/√2

since cos π/4 = 1/√2

cos y = – cos π/4 = cos (π-π/4)

cos y = cos (3π/4)

We, know the range of principal value of cos-1   is [0,π]

Hence, the principal value of  cos-1  (-1/2) is 3π/4

Let y = cosec-1(-√2)

cosec y = -√2

since cosec π/4 = √2

cosec y = -cosec π/4 =cos(-π/4)

We know the range of cosec-1  is [-π/2,π/2]

Therefore the principal value of cosec-1(√2) is -π/4

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