If sec theta + tan theta = 2 + sqrt{5},then t | vedclass

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(A) Given: $\sec 	heta + 	an 	heta = 2 + \sqrt{5} \quad \dots(1)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec

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$\frac{2}{\sqrt{5}}$

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(A) Given: $\sec 	heta + 	an 	heta = 2 + \sqrt{5} \quad \dots(1)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Therefore,$\sec 	heta - 	an 	heta = \frac{1}{\sec 	heta + 	an 	heta} = \frac{1}{2 + \sqrt{5}}$.Rationalizing the denominator: $\sec 	heta - 	an 	heta = \frac{1}{2 + \sqrt{5}} 	imes \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \quad \dots(2)$.Adding equations $(1)$ and $(2)$:$(\sec 	heta + 	an 	heta) + (\sec 	heta - 	an 	heta) = (2 + \sqrt{5}) + (\sqrt{5} - 2)$$2 \sec 	heta = 2 \sqrt{5} \implies \sec 	heta = \sqrt{5}$.Since $\sec 	heta = \frac{	ext{Hypotenuse}}{	ext{Base}} = \frac{\sqrt{5}}{1}$,we have $	ext{Hypotenuse} = \sqrt{5}$ and $	ext{Base} = 1$.Using Pythagoras theorem,$	ext{Perpendicular} = \sqrt{(	ext{Hypotenuse})^2 - (	ext{Base})^2} = \sqrt{(\sqrt{5})^2 - 1^2} = \sqrt{5 - 1} = \sqrt{4} = 2$.Thus,$\sin 	heta = \frac{	ext{Perpendicular}}{	ext{Hypotenuse}} = \frac{2}{\sqrt{5}}$.

If $\sec \theta + \tan \theta = 2 + \sqrt{5}$,then the value of $\sin \theta$ is

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