When there are two values, one is positive and the other is negative such that they are numerically equal, then the principal value of the inverse trigonometric function is the positive one.

**Principal Value of Inverse Trigonometric Functions**

Let us recall that the **principal value **of a inverse
trigonometric function at a point *x *is the value of the inverse function
at the point *x *, which lies in the range of principal branch. For
instance, the principal value of cos^{−1} (√3/2) is π/6. Since π /6 ∈
[0, π].

When there are two values, one is positive and the other is
negative such that they are numerically equal, then the principal value of the
inverse trigonometric function is the positive one. Now, we list out the **principal domain **and **range **of **trigonometric
functions **and
the **domain **and **range **of **inverse trigonometric
functions**.

Find the principal value of

(i) cosec^{-1} (-1) (ii) sec^{-1} (-2) .

(i) Let cosec^{-1} (-1) = *y *. Then, cosec *y *= -1

Since the **range of principal value branch **of y= cosec^{-1} x* *is [- *π/2
*,* π/2*] \ {0} and

Thus, the principal value of cosec^{-1} (-1) is – π/2 .

(ii) Let *y *= sec^{-1} (-2) . Then, sec *y *= -2 .

By definition, the **range of the principal value branch **of *y *= sec^{-1} *x *is [0,π ]\ {π /2} .

Let us find *y *in [0,*π *] – {*π/2*} such that
sec *y *= -2 .

But, sec y = −2 ⇒ cos y = − 1/2 .

Now, cos y =- 1/2 = -cos π/3 = cos (π – π/3 ) = cos 2π/3 .
Therefore, y = 2π/3 .

Since 2π/3 ∈ [0, π ] \ {π/2 } , the
principal value of sec^{-1} (-2) is 2π/3 .

Find the value of sec^{-1}(- 2√3 / 2)

If cot^{-1} ( 1/7 ) = θ , find the value of cos θ .

By definition, cot^{−1} x ∈ (0, π) .

Therefore, cot^{-1} (1/7) = θ implies cot θ ∈ (0,π ) .

But cot^{-1} ( 1/7 ) = θ implies cot θ = 1/7 and hence tan θ = 7 and θ is
acute.

Using tan θ = 7/1 , we construct a right triangle as shown . Then,
we have, cosθ = 1/ 5√2 .

**S**how that , *x *> 1 .

We construct a right triangle with the given data.

From the triangle, secα = x/1 = x . Thus, α = sec^{-1} x .

Tags : Definition, Solved Example Problems , 12th Mathematics : UNIT 4 : Inverse Trigonometric Functions

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12th Mathematics : UNIT 4 : Inverse Trigonometric Functions : Principal Value of Inverse Trigonometric Functions | Definition, Solved Example Problems

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