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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