sec {50^o} + tan {50^o} is equal to | vedclass

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(C) We know that $\sec 	heta + 	an 	heta = 	an(\frac{	heta}{2} + 45^o)$. However,we can simplify this expression using trigonometric identities.L

Vedclass · Accepted Answer

$	an {20^o} + 2	an {50^o}$

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(C) We know that $\sec 	heta + 	an 	heta = 	an(\frac{	heta}{2} + 45^o)$. However,we can simplify this expression using trigonometric identities.Let $x = \sec {50^o} + 	an {50^o} = \frac{1 + \sin {50^o}}{\cos {50^o}}$.Using the identities $1 + \sin 	heta = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$ and $\cos 	heta = \cos^2 \frac{	heta}{2} - \sin^2 \frac{	heta}{2}$,we get:$x = \frac{(\cos {25^o} + \sin {25^o})^2}{(\cos {25^o} - \sin {25^o})(\cos {25^o} + \sin {25^o})} = \frac{\cos {25^o} + \sin {25^o}}{\cos {25^o} - \sin {25^o}} = 	an(45^o + 25^o) = 	an {70^o}$.Now,consider the identity $	an(70^o - 20^o) = 	an {50^o} = \frac{	an {70^o} - 	an {20^o}}{1 + 	an {70^o} 	an {20^o}}$.Since $	an {70^o} = \cot {20^o}$,we have $	an {70^o} 	an {20^o} = 1$.Thus,$	an {50^o} = \frac{	an {70^o} - 	an {20^o}}{1 + 1} = \frac{	an {70^o} - 	an {20^o}}{2}$.$2 	an {50^o} = 	an {70^o} - 	an {20^o}$.$	an {70^o} = 2 	an {50^o} + 	an {20^o}$.Since $\sec {50^o} + 	an {50^o} = 	an {70^o}$,it follows that $\sec {50^o} + 	an {50^o} = 	an {20^o} + 2 	an {50^o}$.

$\sec {50^o} + \tan {50^o}$ is equal to

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