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If $sec^2A + tan^2 A = \frac{4}{17}$, then $sec^4A - tan^A$ is equal to: |
$\frac{13}{17}$ $\frac{4}{13}$ $\frac{4}{17}$ $\frac{5}{17}$ |
$\frac{4}{17}$ |
sec4A - tan4A = ( sec2A - tan2A )× ( sec2A + tan2A ) ---(1) we know , ( sec2A - tan2A ) = 1 & ( sec2A + tan2A ) = \(\frac{4 }{17}\) ( given) Put in equation 1 = 1 × \(\frac{4 }{17}\) = \(\frac{4 }{17}\) |
