If sec alpha + tan alpha = 2, then the value | vedclass

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(C) We know the identity $\sec^{2} \alpha - 	an^{2} \alpha = 1.$This can be factored as $(\sec \alpha + 	an \alpha)(\sec \alpha - 	an \alpha) = 1.$

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) We know the identity $\sec^{2} \alpha - 	an^{2} \alpha = 1.$This can be factored as $(\sec \alpha + 	an \alpha)(\sec \alpha - 	an \alpha) = 1.$Given $\sec \alpha + 	an \alpha = 2,$ we substitute this into the identity:$2(\sec \alpha - 	an \alpha) = 1 \implies \sec \alpha - 	an \alpha = 0.5.$Now,add the two equations:$(\sec \alpha + 	an \alpha) + (\sec \alpha - 	an \alpha) = 2 + 0.5 = 2.5.$$2 \sec \alpha = 2.5 \implies \sec \alpha = 1.25 = 5/4.$Since $\cos \alpha = 1 / \sec \alpha,$ we have $\cos \alpha = 4/5 = 0.8.$Using $\sin^{2} \alpha + \cos^{2} \alpha = 1,$ we find $\sin^{2} \alpha = 1 - (4/5)^{2} = 1 - 16/25 = 9/25.$Since $0 < \alpha < 90^{\circ},$ $\sin \alpha$ must be positive,so $\sin \alpha = \sqrt{9/25} = 3/5 = 0.6.$

If $\sec \alpha + \tan \alpha = 2,$ then the value of $\sin \alpha$ is (assume that $0 < \alpha < 90^{\circ}$).

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