cos[tan-1{sin(cot-1 x)}] = [(1 + x^2)/(1 - x^2)]^-1
rn
prove it!!!!

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cos[tan-1{sin(cot-1 x)}] = [(1 + x^2)/(1 - x^2)]^-1
rn
prove it!!!!

Shridhar Kaushik · Accepted Answer

Taking L H S , $\cos [\tan^{- 1} {\sin (\cot^{- 1} x)}] Let \cot^{- 1} x = θ \Rightarrow x = \cot θ ∴ \cos [\tan^{- 1} {\sin (\cot^{- 1} x)}] = \cos [\tan^{- 1} {\sin θ}] (1) Now, \cot θ = x Also, 1 + \cot^{2} θ = {cosec}^{2} θ \Rightarrow 1 + x^{2} = {cosec}^{2} θ \Rightarrow cosec θ = \sqrt{1 + x^{2}} \Rightarrow \sin θ = \frac{1}{\sqrt{1 + x^{2}}} [∵ \sin θ = \frac{1}{cosec θ}]$ âˆ´ From (1), we get â€“ $\cos [\tan^{- 1} {\sin (\cot^{- 1} x)}] = \cos [\tan^{- 1} (\frac{1}{\sqrt{1 + x^{2}}})] (3) Let \tan^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}) = y \Rightarrow \frac{1}{\sqrt{1 + x^{2}}} = \tan y (4)$ From (3), $\cos [\tan^{- 1} {\sin (\cot^{- 1} x)}] = \cos y (5)$ Our aim is to find the value of cos y $Since, \tan y = \frac{1}{\sqrt{1 + x^{2}}} and \sec^{2} y = 1 + \tan^{2} y = 1 + \frac{1}{1 + x^{2}} \Rightarrow \sec^{2} y = \frac{2 + x^{2}}{1 + x^{2}} \Rightarrow \cos^{2} y = \frac{1 + x^{2}}{2 + x^{2}} \Rightarrow \cos y = \sqrt{\frac{1 + x^{2}}{2 + x^{2}}}$ âˆ´ From (5), $\cos [\tan^{- 1} {\sin (\cot^{- 1} x)}] = \sqrt{\frac{1 + x^{2}}{2 + x^{2}}}$ Hence, the R H S given by you seems to be incorrect The value on the R H S in the statement should be $\sqrt{\frac{1 + x^{2}}{2 + x^{2}}}$ Please check it again and then do get back to us

cos[tan-1{sin(cot-1 x)}] = [(1 + x^2)/(1 - x^2)]^-1rnprove it!!!!

cos[tan-1{sin(cot-1 x)}] = [(1 + x^2)/(1 - x^2)]^-1
rn
prove it!!!!