If 2 sec 2 α – sec 4 α – 2 cosec 2 α + cosec 4 α = 15/4, then tan α can be
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(a,d)
⇒ 2(sec 2 α – cosec 2 α ) + (cosec 2 α + sec 2 α ) (cosec 2 α – sec 2 α ) = 
⇒ (cosec 2 α – sec 2 α ) [cosec 2 α + sec 2 α – 2] = 
⇒ 4(cot 2 α – tan 2 α ) (cot 2 α + tan 2 α ) = 15 ⇒ 4(cot 2 α – tan 2 α ) = 15
⇒ 4(1 – tan 8 α ) = 15 tan 4 α ⇒ 4 tan 8 α + 15 tan 4 α – 4 = 0
⇒ 4 tan 8 α + 16 tan 4 α – tan 4 α – 4 = 0 ⇒ (4 tan 4 α – 1) (tan 4 α + 4) = 0
⇒ tan4 α = 
or tan 4 α = – 4 (not possible) ⇒ tan 2 α = ± 
⇒ tan 2 α = + 
⇒ tan α = ± 
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