Given 1−cotAtanA+1−tanAcotA
=cosA(sinA−cosA)sin2A+sinA(cosA−sinA)cos2A
=sinAcosA(sinA−cosA)sin3A−cos3A
=sinAcosAsin2A+cos2A+sinAcosA
=secAcosecA+1
JEE Main 2013 — Mathematics Trigonometry
The expression 1−cotAtanA+1−tanAcotA can be written as :
Held on 7 Apr 2013 · Verified 6 Jul 2026.
tanA+cotA
secA+cosecA
sinAcosA+1
secAcosecA+1
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