Question

`[sin (tan^-1  3/4)]^2 + [sin(tan^-1  4/3)]^2 = ?`

पर्याय

• 0

• 1

• 5

• -1

Answer 1

1

Explanation:

We have,

`[sin (tan^-1  3/4)]^2 + [sin(tan^-1  4/3)]^2`

Let `tan^-1  3/4 = theta => tan theta = 3/4 => sin theta = 3/5`

and `tan^-1  4/3 = phi => tan phi = 4/3 => sin phi = 4/5`

∴ `[sin (tan^-1  3/4)]^2 + [sin (tan^-1  4/3)]^2`

`= (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1`