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\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right) |
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Answer» tan⁻¹1/4 + tan⁻¹ 2/9 = tan⁻¹( 1/4 + 2/9)/(1-(1/4)*(2/9))= tan⁻¹ (1/2) let tan⁻¹(1/2)=θ So, tanθ = 1/2So, we can find cos θ = 2/√5cos 2θ = 2cos²θ -1 = 2*4/5 - 1 = 3/5So, cos 2θ = 3/5hence, 2θ = cos ⁻¹ 3/5 θ = 1/2 cos⁻¹ 3/5but, θ = tan⁻¹1/2 So, tan⁻¹ 1/2 = 1/2 cos⁻¹ 3/5 |
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