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





11 and 12 Grade Math

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