cosA,sinA,cotA are in GP, then tan6A−tan2A is equal to
A
-1
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B
0
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C
1
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D
2
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Solution
The correct option is C 1 As given , sin2A=cos2AsinA So we get sin3A=cos2A Now tan6A−tan2A=sin6Acos6A−tan2A Putting sin6A=cos4A we get tan6A−tan2A=cos4Acos6A−tan2A =sec2A−tan2A=1