MATH CHAPTER 2 INVERSE TRIGNOMETRIC FORMULAS
CH-2 INVERSE TRIGNOMETRIC FORMULA
Let tan(π/4) = 1
then (π/4) = 1/tan
= tan⁻¹(1)
So inverse of a function is
the angle.
in general if tany = x
then y = tan⁻¹x
Similarly sin⁻¹x, cos⁻¹x etc.
Some formulas
1. sin⁻¹(sinx) = x
2. cos⁻¹(cosx) = x
3. tan⁻¹(tanx) = x
4. cot⁻¹(cotx) = x
5. sec⁻¹(secx) = x
6. cosec⁻¹(cosec x) = x
Example: find sin⁻¹(1/2)
Sol. We have to cut sin⁻¹ By
Putting (1/2) in the terms of
Sine
: Sin⁻¹(½) = sin⁻¹(sin 𝝅/6)
= 𝝅/6
Some New formulas
1. Sin⁻¹x + cos⁻¹x = π/2
2. tan⁻¹x + cot⁻¹x = π/2
3. sec⁻¹x + cosec⁻¹x = π/2
Ex -> Find x if
sin⁻¹x + cos⁻¹(1/5) = π/2
Sol.
we have sin⁻¹x = π/2 - cos⁻¹(1/5)
sin⁻¹x = sin⁻¹(1/5) By I
x = 1/5
Ex find sin (tan⁻¹x + cot⁻¹x).
Sol.
we have tan⁻¹x + cot⁻¹x = π/2
∴ sin(π/2) = 1
Same formula.
1. tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy))
2. tan⁻¹x - tan⁻¹y = tan⁻¹((x-y)/(1+xy))
3. sin⁻¹x + sin⁻¹y = sin⁻¹[x√(1-y²) + y√(1-x²)]
4. cos⁻¹x + cos⁻¹y = cos⁻¹[xy - √(1-x²)√(1-y²)]
5. 2tan⁻¹x = tan⁻¹(2x/(1-x²))
2tan⁻¹x = sin⁻¹(2x/(1+x²))
2tan⁻¹x = cos⁻¹((1-x²)/(1+x²))
Sol.
tan⁻¹½ + tan⁻¹⅓
= tan⁻¹( ½+⅓ ) /(1-½×⅓) By I
= tan⁻¹[( 3+2 )/6]/[(6-1)/5]
= tan⁻¹(⁵/₅) = tan⁻¹(1) = tan⁻¹(tanπ/₄) = π/₄
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