Prove that 1/cosa+cos3a + 1/cosa+cos5a + 1/cosa+cos7a + ......+ 1/cosa+cos(2n+1)a = (1/2 )coseca [tan(n+1)a – tana]

Prove that  1/cosa+cos3a + 1/cosa+cos5a + 1/cosa+cos7a + ......+ 1/cosa+cos(2n+1)a = (1/2 )coseca [tan(n+1)a – tana]

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Sol.    Now,  1/cosa+cos(2n+1)a   = 1/2 cos(n+1)acosna
= 1/2sina [sin(n+1-n)a/cos(n+1)acosna]
= 1/2sina[{sin(n+1)acosna-cos(n+1)asinna}/cos(n+1)acosna]
=1/2sina[tan(n+1)a-tanna]
Therefore
1/cosa+cos3a + 1/cosa+cos5a + 1/cosa+cos7a + ......+ 1/cosa+cos(2n+1)a 
=1/2sina[tan2a-tana+tan3a-tan2a+tan4a-tan3a+......+tan(n+1)a-tanna]
=1/2sina[tan(n+1)a-tana]
=(1/2 )coseca [tan(n+1)a – tana]



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