Math, asked by hinanshu39, 1 year ago

cot inverse (cosx-sinx/cosx+sinx)​

Answers

Answered by OmBacchuwar
10

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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