Dear Student, Solution) 24) costan-1x=sincot-134 1As we know that sin θ= cos 90-θ so, sincot-134=cos π2-cot-134Now we know that tan-1x+cot-1x=π2 soπ2-cot-134=tan-134 putting this we get, sincot-134=cos π2-cot-134=costan-134 Now putting this value in equation 1, we getcostan-1x=costan-134 or tan-1x=tan-134or x = 34 Regards! â€‹â€‹â€‹â€‹â€‹â€‹ â€‹â€‹â€‹â€‹â€‹

Plz slvr ques. 24

Dear Student,
Solution) 24)

\cos (\tan^{- 1} x) = \sin (\cot^{- 1} (\frac{3}{4})) . . . (1) As we know that \sin θ = \cos (90 - θ) so, \sin (\cot^{- 1} \frac{3}{4}) = \cos (\frac{π}{2} - \cot^{- 1} (\frac{3}{4})) Now we know that \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2} so \frac{π}{2} - \cot^{- 1} (\frac{3}{4}) = \tan^{- 1} (\frac{3}{4}) putting this we get, \sin (\cot^{- 1} \frac{3}{4}) = \cos (\frac{π}{2} - \cot^{- 1} (\frac{3}{4})) = \cos [\tan^{- 1} (\frac{3}{4})] Now putting this value in equation (1), we get \cos (\tan^{- 1} x) = \cos [\tan^{- 1} (\frac{3}{4})] or \tan^{- 1} x = \tan^{- 1} (\frac{3}{4}) or x = (\frac{3}{4})

Regards!

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