To prove: sin−1(

4

5

)+sin−1(

5

13

) + sin−1(

16

65

) =

π

2

Let sin−1(

4

5

) = x
⇒ sin x =

4

5

⇒ cos x = √1−sin2x =

3

5

sin−1(

5

13

) = y
⇒ sin y =

5

13

⇒ cos y = √1−sin2y =

12

13

sin−1(

16

65

) = z
⇒ sin z =

16

65

⇒ cos z = √1−sin2x =

63

65

tan x =

4

3

, tan y =

5

12

, tan z =

16

63

tan z =

16

63

⇒ cot z =

63

16

... (1)
tan (x + y) =

tan(x+y)

1−tanx.tany

⇒ tan (x + y) =

4 3 + 5 12

1− 20 36

⇒ tan (x + y) =

63

16

⇒ tan (x + y) = cot z ... [from equation (1)]
⇒ tan (x + y) = tan (

π

2

−z)
⇒ x + y =

π

2

- z
⇒ x + y + z =

π

2

∴ sin−1(

4

5

)+sin−1(

5

13

) + sin−1(

16

65

) =

π

2

OR
tan−13x+tan−12x =

π

4

⇒ tan−1(

5x

1−6x2

) =

π

4

. 3x × 2x < 1
⇒ tan [tan−1(

5x

1−6x2

)] = tan

π

4

⇒

5x

1−6x2

= 1
⇒ 1 - 6x2 = 5x
⇒ 6x2 + 5x - 1 = 0
⇒ 6x2 + 6x - x - 1 = 0
⇒ x = - 1 or

1

6

Here (- 3) × (- 2) ≮ 1 [Since (- 3) × (- 2) = 6 >1]
Therefore, x = 1 is not the solution.
When substituting x =

1

6

in 3x 2x, we have,
3 ×

1

6

× 2 ×

1

6

=

1

2

×

1

3

=

1

6

< 1
Hence x =

1

6

is the solution of the given equation