cot inverse ( cosx-sinx/coxx+sinx)
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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
= 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
rakhithakur:
hey is it right
third step smaj nhi aaya
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the answer I s the RHS
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