cot inverse (cosx-sinx/cosx+sinx)
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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)]
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hinanshu39:
right answer
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