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

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(C) We know that $	an {50^o} = 	an ({70^o} - {20^o}) = \frac{	an {70^o} - 	an {20^o}}{1 + 	an {70^o} 	an {20^o}}$.Since $	an {70^o} = \cot {20^

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

Vedclass · Answer

(C) We know that $	an {50^o} = 	an ({70^o} - {20^o}) = \frac{	an {70^o} - 	an {20^o}}{1 + 	an {70^o} 	an {20^o}}$.Since $	an {70^o} = \cot {20^o} = \frac{1}{	an {20^o}}$,we have $	an {70^o} 	an {20^o} = 1$.Substituting this into the equation: $	an {50^o} = \frac{	an {70^o} - 	an {20^o}}{1 + 1} = \frac{	an {70^o} - 	an {20^o}}{2}$.Thus,$2 	an {50^o} = 	an {70^o} - 	an {20^o}$,which implies $	an {70^o} = 2 	an {50^o} + 	an {20^o}$.Also,$\sec {50^o} + 	an {50^o} = \frac{1 + \sin {50^o}}{\cos {50^o}} = \frac{\cos {40^o} + \sin {40^o}}{\sin {40^o}} = \cot {40^o} + 1 = 	an {50^o} + 1$ (This is not the direct path).Using the identity $	an {70^o} = \sec {50^o} + 	an {50^o}$:Since $	an {70^o} = 	an (90^o - 20^o) = \cot {20^o} = \frac{\cos {20^o}}{\sin {20^o}} = \frac{1 + \cos {40^o}}{\sin {40^o}} = \frac{2 \cos^2 {20^o}}{2 \sin {20^o} \cos {20^o}} = \cot {20^o}$.Actually,$\sec {50^o} + 	an {50^o} = 	an (45^o + 25^o) = 	an {70^o} = 2 	an {50^o} + 	an {20^o}$ is incorrect. Correct derivation: $\sec {50^o} + 	an {50^o} = 	an (45^o + 25^o)$. Given the options,the correct identity is $\sec {50^o} + 	an {50^o} = 	an {20^o} + 2 	an {50^o}$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

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

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