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\(\frac{d}{dx}[tan^{-1}(\frac{cosx+sinx}{cosx-sinx})]\) = d/dx[tan-1((cosx+sinx)/(cosx-sinx))] =(A) \(\frac{\pi}{4}\) + x(B) 1(C) -1(D) tan\(\frac{x}{2}\) |
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Answer» Option : (B) 1 |
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