if m=cosA-sinA and n=cosA-sinA show that √m/√n+√n/√m=2/√1-tan^2A
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=Given, m=cosA
=−sinAn=cosA+sinASo mn−nm=m2−
=n2mn=(cosA−sinA)2−(cosA+sinA)2cos2A
=−sin2A=cos2A+sin2A−2sinAcosA−cos2A=−sin2A
==−2sinAcosAcos2A−sin2A=−4sinAcosAcos2A
=−sin2A=−4sinAcosAsinAcosAcos2AsinAcosA−
=sin2AsinAcosA=−4cotA−tanA
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=Given, m=cosA
=−sinAn=cosA+sinASo mn−nm=m2−
=n2mn=(cosA−sinA)2−(cosA+sinA)2cos2A
=−sin2A=cos2A+sin2A−2sinAcosA−cos2A=−sin2A
==−2sinAcosAcos2A−sin2A=−4sinAcosAcos2A
=−sin2A=−4sinAcosAsinAcosAcos2AsinAcosA−
=sin2AsinAcosA=−4cotA−tanA
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mayank042:
your answer is not understandable
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