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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^

Vedclass · Accepted Answer

$	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 $\sec {50^o} + 	an {50^o} = 	an {20^o} + 2 	an {50^o}$.

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

Similar Questions

If $\tan A + \cot A = 4$,then $\tan^4 A + \cot^4 A$ is equal to

If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then
$(A) \tan ^2 x=\frac{2}{3}$ $(B) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C) \tan ^2 x=\frac{1}{3}$ $(D) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

$\operatorname{sech}^{-1}(\sin \theta)$ is equal to

Observe that,at any instant,the minute and hour hands of a clock make two angles between them whose sum is $360^{\circ}$. At $6:15$,the difference between these two angles is $....^{\circ}$.

The number of all possible integral values of $n > 2$ such that $\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module