Chapter 2 Inverse Trigonometric Functions
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See all →- cos-1(cos\(\frac { 13π }{ 6 }\)) cos⁻¹(cos 13π / 6 ) = cos⁻¹cos(2π + π / 6 = cos⁻¹cos π / 6 ) = π / 6 1
- tan-1(tan\(\frac { 7π }{ 6 }\)) tan⁻¹(tan 7π / 6 ) = tan⁻¹tan(π + π / 6 = tan⁻¹tan π / 6 ) = π / 6 . 1
- tan-1\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos-1\(\frac { 1-x }{ 1+x }\), x ∈ [0, 1] Put x = tan²θ. ∴ θ = tan⁻¹ √(x) R.H.S. = 1 / 2 cos⁻¹ 1-x / 1+x = 1 / 2 cos⁻¹ (frac1-tan ² θ1+tan ² θ) = 1 / 2 cos⁻¹(cos2θ) = 1 / 2 x 2θ = θ.... 1
- 2tan-1(cos x) = tan-1(cosec x) Now L.H.S = 2tan⁻¹(cos x) = tan⁻¹ (frac2 cos x1-cos ² x) = tan⁻¹ (frac2 cos xsin ² x) Putting this value in the given equation, we get tan⁻¹... 1
- sin tan-1x, |x| < 1 is equal to (A) \(\frac{x}{\sqrt{1-x^{2}}}\) (B) \(\frac{1}{\sqrt{1-x^{2}}}\) Let tan⁻¹x = α ∴ tan α = x. So, sin α = fracxsqrt1+x² or α = sin⁻¹ fracxsqrt1+x² Now sin tan⁻¹x = sin α = sin(sin⁻¹ fracxsqrt1+x² ) = fracxs... 1
- If sin-1(1-x)-2 sin-1x = \(\frac { π }{ 2 }\), then x is equal to (A) 0, \(\frac { 1 }{ 2 }\) sin⁻¹(1-x)-2 sin⁻¹x = π / 2 Putting π / 2 = sin⁻¹(1 – x) + cos⁻¹(1 – x), sin⁻¹(1 – x) – 2 sin⁻¹ = sin⁻¹(1 – x) + cos⁻¹(1 – x) ⇒ – 2 sin⁻¹x =... 1