If sec theta - tan theta = frac{1}{sqrt{3}},t | vedclass

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Question

(A) We know the identity $\sec^2 	heta - 	an^2 	heta = 1$.This can be factored as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Give

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$\frac{2}{3}$

Vedclass · Answer

(A) We know the identity $\sec^2 	heta - 	an^2 	heta = 1$.This can be factored as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Given $\sec 	heta - 	an 	heta = \frac{1}{\sqrt{3}}$,we substitute this into the identity:$\frac{1}{\sqrt{3}}(\sec 	heta + 	an 	heta) = 1 \implies \sec 	heta + 	an 	heta = \sqrt{3}$.Now,we have a system of two equations:$1) \sec 	heta - 	an 	heta = \frac{1}{\sqrt{3}}$$2) \sec 	heta + 	an 	heta = \sqrt{3}$Adding the two equations: $2 \sec 	heta = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1+3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \implies \sec 	heta = \frac{2}{\sqrt{3}}$.Subtracting the first from the second: $2 	an 	heta = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \implies 	an 	heta = \frac{1}{\sqrt{3}}$.Therefore,$\sec 	heta \cdot 	an 	heta = \left(\frac{2}{\sqrt{3}}ight) \cdot \left(\frac{1}{\sqrt{3}}ight) = \frac{2}{3}$.

If $\sec \theta - \tan \theta = \frac{1}{\sqrt{3}}$,the value of $\sec \theta \cdot \tan \theta$ is

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