Question 4: Find the principal value of tan¯¹ (− √3)
Class 12 - Math - Inverse Trigonometric Functions
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Let tan-¹(-√3) = ∅
where ∅ is the principal branch of tan-¹(-√3)
therefore tan∅ = -√3
since the range of principal value of branch of tan-¹x is [+π/2,π/2]. therefore -π/2≤∅≤π/2
Now tan∅ = -√3 = -tan(π/3) = tan(-π/3)
thus ∅ = -π/3 as tan-¹(-π/3) = -√3 and
-π/3 € [-π/2,π/2]
where ∅ is the principal branch of tan-¹(-√3)
therefore tan∅ = -√3
since the range of principal value of branch of tan-¹x is [+π/2,π/2]. therefore -π/2≤∅≤π/2
Now tan∅ = -√3 = -tan(π/3) = tan(-π/3)
thus ∅ = -π/3 as tan-¹(-π/3) = -√3 and
-π/3 € [-π/2,π/2]
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