# Problems on Inverse Trigonometric Function

We will solve different types of problems on inverse trigonometric function.

1. Find the values of sin (cos$$^{-1}$$ 3/5)

Solution:

Let, cos$$^{-1}$$ 3/5 = θ

Therefore, cos θ = 3/5

Therefore, sin θ = √(1 - cos$$^{2}$$  θ) = √(1 - 9/25) = √(16/25) = 4/5 .

Therefore, sin (cos$$^{-1}$$ 3/5) = sin θ = 4/5.

2. Find the values of tan$$^{-1}$$ sin (- π/2)

Solution:

tan$$^{-1}$$ sin (- π/2)

= tan$$^{-1}$$ (- sin π/2)

= tan$$^{-1}$$ (- 1), [Since - sin π/2 = -1]

= tan$$^{-1}$$(- tan π/4), [Since tan π/4 = 1]

= tan$$^{-1}$$ tan (-π/4)

= - π/4.

Therefore, tan$$^{-1}$$ sin (- π/2) = - π/4

3. Evaluate: sin$$^{-1}$$ (sin 10)

Solution:

We know that sin$$^{-1}$$ (sin θ) = θ, if - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$.

Here, θ = 10 radians which does not lie between - $$\frac{π}{2}$$ and $$\frac{π}{2}$$. But 3π - θ i.e., 3π - 10 lies between - $$\frac{π}{2}$$ and $$\frac{π}{2}$$ and sin (3π - 10) = sin 10.

Now, sin$$^{-1}$$ (sin 10)

= sin^-1 (sin (3π - 10)

= 3π - 10

Therefore, sin$$^{-1}$$ (sin 10) = 3π - 10.

4. Find the values of cos (tan$$^{-1}$$ ¾)

Solution:

Let, tan$$^{-1}$$ ¾ = θ

Therefore, tan θ = ¾

We know that sec$$^{2}$$ θ - tan$$^{2}$$ θ = 1

⇒ sec θ = √(1 + tan$$^{2}$$ θ)

⇒ sec θ = √(1 + (3/4)$$^{2}$$)

⇒ sec θ = √(1 + 9/16)

⇒ sec θ = √(25/16)

⇒ sec θ = 5/4

Therefore, cos θ = 4/5

⇒ θ = cos$$^{-1}$$ 4/5

Now, cos (tan$$^{-1}$$ ¾) = cos (cos$$^{-1}$$ 4/5) = 4/5

Therefore, cos (tan$$^{-1}$$ ¾) = 4/5

5. Find the values of sec csc$$^{-1}$$ (2/√3)

Solution:

sec csc$$^{-1}$$ (2/√3)

= sec csc$$^{-1}$$ (csc π/3)

= sec (csc$$^{-1}$$csc π/3)

= sec π/3

= 2

Therefore, sec csc$$^{-1}$$ (2/√3) = 2

Inverse Trigonometric Functions

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