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Ask iit jee aieee pet cbse icse state board experts Expert Question: trignometry
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johncena_raw07 (0)

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(cotA*cosA)/cotA+cosA=(cotA-cosA)/cotA*cosA

    
anchitsaini (4240)

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JEE and OLYMPIA INFINATUM


http://iit-redefined.theforum.name/index.php


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edison (3966)

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To Prove (cotA*cosA)/cotA+cosA=(cotA-cosA)/cotA*cosA



LHS = (cotA*cosA)/cotA+cosA



Rationalising this we have



LHS = (cotA*cosA) (cotA-cosA) / (cotA+cosA) (cotA-cosA)



= (cotA*cosA) (cotA-cosA) / (cot2A - cos2 A)



BY substituting cot A = cosA/ SinA we have



LHS = cos2A sin A(cotA-cosA) / (cos2 A - sin2 A cos2 A)



= sin A(cotA-cosA) / cos2 A



Dividng Numerator and dinominator by Sin A we obtain



LHS = (cotA-cosA)/cotA*cosA = RHS







 


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