Question

Tg(arcsin 1 - arccos корень 2/2)​

Answer 1

$tg(arcSin1-arcCos\frac{\sqrt{2}}{2})=tg(\frac{\pi }{2}-\frac{\pi }{4})=tg\frac{2\pi-\pi}{4}=tg\frac{\pi}{4}=1\\\\Otvet:\boxed{1}$

Answer 2

To evaluate the expression Tg(arcsin(1) - arccos(√2/2)), let's break it down step by step:

1. First, let's evaluate the values inside the trigonometric functions:

   arcsin(1) = π/2 arccos(√2/2) = π/4

2. Now, substitute these values back into the expression:

   Tg(π/2 - π/4)

3. Simplify the expression within the tangent function:

   π/2 - π/4 = π/4

4. Finally, calculate the tangent of π/4:

   Tg(π/4) = tan(π/4) = 1

Therefore, Tg(arcsin(1) - arccos(√2/2)) equals 1.