If sec A = a + left(frac{1}{4a}right),then se | vedclass

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Question

(A) We know the identity $	an^2 A = \sec^2 A - 1$.Substituting the given value of $\sec A$:$	an^2 A = \left(a + \frac{1}{4a}ight)^2 - 1$$= a^2 + 2

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$2a$ or $\frac{1}{2a}$

Vedclass · Answer

(A) We know the identity $	an^2 A = \sec^2 A - 1$.Substituting the given value of $\sec A$:$	an^2 A = \left(a + \frac{1}{4a}ight)^2 - 1$$= a^2 + 2(a)\left(\frac{1}{4a}ight) + \left(\frac{1}{4a}ight)^2 - 1$$= a^2 + \frac{1}{2} + \frac{1}{16a^2} - 1$$= a^2 - \frac{1}{2} + \frac{1}{16a^2} = \left(a - \frac{1}{4a}ight)^2$Therefore,$	an A = \pm \left(a - \frac{1}{4a}ight)$.Case $1$: $	an A = a - \frac{1}{4a}$$\sec A + 	an A = \left(a + \frac{1}{4a}ight) + \left(a - \frac{1}{4a}ight) = 2a$.Case $2$: $	an A = -\left(a - \frac{1}{4a}ight) = -a + \frac{1}{4a}$$\sec A + 	an A = \left(a + \frac{1}{4a}ight) + \left(-a + \frac{1}{4a}ight) = \frac{2}{4a} = \frac{1}{2a}$.Thus,the value is $2a$ or $\frac{1}{2a}$.

If $\sec A = a + \left(\frac{1}{4a}\right)$,then $\sec A + \tan A =$

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