If $sec^2A + tan^2 A = \frac{4}{17}$, then $sec^4A - tan^A$ is equal to:
Answer & explanation
Correct answer: option 3
sec4A - tan4A
= ( sec2A - tan2A )× ( sec2A + tan2A ) ---(1)
we know , ( sec2A - tan2A ) = 1
& ( sec2A + tan2A ) = \(\frac{4 }{17}\) ( given)
Put in equation 1
= 1 × \(\frac{4 }{17}\)
= \(\frac{4 }{17}\)