Answer:
(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
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