If sec theta = frac{5}{3},then tan theta = ld | vedclass

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(C) Given that $\sec 	heta = \frac{5}{3}$.We use the trigonometric identity $	an^2 	heta = \sec^2 	heta - 1$.Substituting the value of $\sec 	het

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$\frac{4}{3}$

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(C) Given that $\sec 	heta = \frac{5}{3}$.We use the trigonometric identity $	an^2 	heta = \sec^2 	heta - 1$.Substituting the value of $\sec 	heta$:$	an^2 	heta = \left(\frac{5}{3}ight)^2 - 1$$	an^2 	heta = \frac{25}{9} - 1$$	an^2 	heta = \frac{25 - 9}{9} = \frac{16}{9}$Taking the square root on both sides:$	an 	heta = \sqrt{\frac{16}{9}} = \frac{4}{3}$.

If $\sec \theta = \frac{5}{3}$,then $\tan \theta = \ldots$

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