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(1+cotA+tanA)(sinA-cosA) =sinAtanA-cotAcosA

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(1+cotA+tanA)(sinA-cosA) =sinAtanA-cotAcosA
  • 2 answers

Tera Baap 6 years, 10 months ago

Taking l.h.s. (1+cosA/sinA+sinA/cosA)(sinA-cosA) ,=(sinA.cosA+sin sq A+cos sq A)/sinA.cosA-(sinA-cosA)=1+sinA.cosA/sinA.cosA (sinA-cosA)=(cosecA.secA+1)(sinA-cosA) similarly prove r.h.s into sinA&cosA proving bahut long hai.

Gaurav Seth 6 years, 10 months ago

(1+cotA+tanA)(sinA-cosA)
= (sinA+sinAcotA +tanAsinA-cosA -cosAcotA -tanAcosA)
=(sinA+sinAcosA/sinA+tanAsinA-cosA-cosAcotA -cosAsinA/cosA)
= (sinA+cosA+tanAsinA -cosA- cosAcotA -sinA)
= tanAsinA -cosAcotA 
= sinAtanA-cotAcosA

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