If sec theta = 1frac{1}{4},then tan frac{thet | vedclass

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(A) Given that $\sec 	heta = 1\frac{1}{4} = \frac{5}{4}$.We know the identity $\sec 	heta = \frac{1 + 	an^2(	heta/2)}{1 - 	an^2(	heta/2)}$.Subst

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$\frac{1}{3}$

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(A) Given that $\sec 	heta = 1\frac{1}{4} = \frac{5}{4}$.We know the identity $\sec 	heta = \frac{1 + 	an^2(	heta/2)}{1 - 	an^2(	heta/2)}$.Substituting the value of $\sec 	heta$:$\frac{5}{4} = \frac{1 + 	an^2(	heta/2)}{1 - 	an^2(	heta/2)}$.Cross-multiplying gives:$5(1 - 	an^2(	heta/2)) = 4(1 + 	an^2(	heta/2))$.$5 - 5	an^2(	heta/2) = 4 + 4	an^2(	heta/2)$.$5 - 4 = 4	an^2(	heta/2) + 5	an^2(	heta/2)$.$1 = 9	an^2(	heta/2)$.$	an^2(	heta/2) = \frac{1}{9}$.Taking the square root,$	an(	heta/2) = \frac{1}{3}$.

If $\sec \theta = 1\frac{1}{4}$,then $\tan \frac{\theta }{2} = $

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