Question

prove that cot inverse[cos x+sin x/cos x -sin x]= pi/4 -x

Answer 1

LHS = cot⁻¹[ (cosx + sinx)/(cosx - sinx)]
= Cot⁻¹[(cosx/sinx + sinx/sinx)/(cosx/sinx - sinx/sinx)]
= Cot⁻¹[cotx + 1)/(cotx - 1)]
= Cot⁻¹[ (cotx.cotπ/4 + 1)/(cotx - cotπ/4)]
= Cot⁻¹ [ cot(π/4 - x)]
= π/4 - x = RHS

[ Note :- cot(A - B) = (cotA.cotB +1)/(cotB - cotA)]