Question

If sec^2α-sec^4α-2cosec^2α+cosec^4α=15/4.
then what's the value of tan^2α.

Answer 1

Answer:

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Given : 2sec2α – sec4α – 2cosec2α + cosec4α = (15/4)

consider LHS = 2(1 + tan2α) – (1 + tan2α)2 – 2(1 + cot2α) + (1 + cot2α)2     

  = 2[1 + tan2α – 1 – cot2α] – (1 + 2tan2α + tan4α) + (1 + 2cot2α + cot4α)

= 2(tan2α – cot2α) – 2tan2α – tan4α + 2cot2α + cot4α

= cot4α – tan4α    

= [1 / (tan4α)] – tan4α   

∴ [{1 – (tan8α)}/(tan4α)] = (15/4)    

Let tan4α = P 

Hence, [(1 – P2)/P] = (15/4) 

 ∴ 4P2 + 15P – 4 = 0  

Hence, P= -4 or ( 1/4)

i.e tan^4 a = (1/4)

So, tan^2 a = 1/2

Hope it's helpful...... ☺