Plsss me simplify guys plz don't delete my questions

(i) sin⁻¹ [ x / a ]

(iii) tan⁻¹ [ tan(π/4 − x) ] = π/4 − x + kπ {-π/4 + kπ ≤ x ≤ 3π/4 + kπ}

(v) tan⁻¹ [ √(1 − x²) / √(1 + x²) ] + π/4

Step-by-step explanation:

(i) tan⁻¹ [ x / √(a² − x²) ]

Draw a right triangle, where x is one of the sides and a is the hypotenuse.  Using Pythagorean theorem, the other side is √(a² − x²).  If we say the angle opposite of the x side is θ, then:

θ = tan⁻¹ [ x / √(a² − x²) ]

But we can also define θ using sine:

θ = sin⁻¹ [ x / a ]

(iii) tan⁻¹ [ (cos x − sin x) / (cos x + sin x) ]

Factor cos x from numerator and denominator:

tan⁻¹ [ cos x (1 − tan x) / (cos x (1 + tan x)) ]

tan⁻¹ [ (1 − tan x) / (1 + tan x) ]

Using angle sum formula:

tan⁻¹ [ 1 / tan(x + π/4) ]

tan⁻¹ [ cot(x + π/4) ]

Using phase shift:

tan⁻¹ [ tan(π/2 − (x + π/4)) ]

tan⁻¹ [ tan(π/2 − x − π/4) ]

tan⁻¹ [ tan(π/4 − x) ]

Or:

π/4 − x + kπ {-π/4 + kπ ≤ x ≤ 3π/4 + kπ}

(v) tan⁻¹ [ (√(1 + x²) + √(1 − x²)) / (√(1 + x²) − √(1 − x²)) ]

Notice that (√(1 + x²))² + (√(1 − x²))² = 2.

Therefore, we can substitute √(1 + x²) = √2 cos u, and √(1 − x²) = √2 sin u.

tan⁻¹ [ (√2 cos u + √2 sin u) / (√2 cos u − √2 sin u) ]

tan⁻¹ [ (cos u + sin u) / (cos u − sin u) ]

Notice this is the inverse of what we had in problem (iii).  Therefore, using the same logic:

tan⁻¹ [ tan(u + π/4) ]

u + π/4

Substituting back:

= sin⁻¹ [ √(1 − x²) / √2 ] + π/4

Or:

= cos⁻¹ [ √(1 + x²) / √2 ] + π/4

Or:

= tan⁻¹ [ √(1 − x²) / √(1 + x²) ] + π/4

5/12, or 5: 12

step-by-step explanation:

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer