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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