d/dx[tan^(-1)((cosx+sinx)/(cosx-sinx))] =

\(\frac{d}{dx}[tan^{-1}(\frac{cosx+sinx}{cosx-sinx})]\) =

d/dx[tan^-1((cosx+sinx)/(cosx-sinx))] =

(A) \(\frac{\pi}{4}\) + x

(B) 1

(C) -1

(D) tan\(\frac{x}{2}\)

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

Salim Ahmed Kawsar

Option : (B) 1

4 views Answered 2023-02-05 15:29

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