If sec theta - tan theta = frac{a+1}{a-1}, th | vedclass

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(B) We know that $\sec^{2} 	heta - 	an^{2} 	heta = 1.$This can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1.$Given $\s

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$\frac{a^{2}-1}{a^{2}+1}$

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(B) We know that $\sec^{2} 	heta - 	an^{2} 	heta = 1.$This can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1.$Given $\sec 	heta - 	an 	heta = \frac{a+1}{a-1}.$Therefore,$\sec 	heta + 	an 	heta = \frac{1}{\sec 	heta - 	an 	heta} = \frac{a-1}{a+1}.$Adding the two equations: $(\sec 	heta - 	an 	heta) + (\sec 	heta + 	an 	heta) = \frac{a+1}{a-1} + \frac{a-1}{a+1}.$$2 \sec 	heta = \frac{(a+1)^{2} + (a-1)^{2}}{(a-1)(a+1)} = \frac{(a^{2} + 2a + 1) + (a^{2} - 2a + 1)}{a^{2}-1} = \frac{2(a^{2}+1)}{a^{2}-1}.$Thus,$\sec 	heta = \frac{a^{2}+1}{a^{2}-1}.$Since $\cos 	heta = \frac{1}{\sec 	heta},$ we get $\cos 	heta = \frac{a^{2}-1}{a^{2}+1}.$

If $\sec \theta - \tan \theta = \frac{a+1}{a-1},$ then $\cos \theta =$

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