Question
Tue May 01, 2012 By:
RESPECTED TOPPER LEARNING FACULTY,

I HEREBY BESEECH U TO PROVE THE ABOVE URGENTLY AS I HAVE TO GIVE A TEST RIGHT NOW.

tan2A - tanA = tan A sec 2A cot2A - tanA = cos3A/cosA.sin2A

Expert Reply
Tue May 01, 2012
These are two different questions.
1) tan 2A - tan A = tan A sec2A
 
LHS = tan 2A - tan A
= (2tan A/1 - tan2A) - tan A
= 2tan A - tan A + tan3A/1 - tan2A
= tan A + tan3A/1 - tan2A
= sinA / cosA(cos2A - sin2A)      (by changing tan A as sin A / cosA)
= sinA / cosA. cos2A
= tanA sec2A = RHS
 
2) cot2A - tanA = cos3A/cosA.sin2A
 
LHS = 1/ (tan2A) - tanA
= (1 - tan2A)/ (2tanA) - tanA
= (1 - 3tan2A)/ 2 tanA
= (cos2A - 3sin2A)/ 2sin A cosA     (by changing tan A as sin A / cosA)
= (4 cos2A - 3)/ sin2A
= cos3A/ cosA. sin2A = RHS
 
Hence proved.
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