Question

seca+tana=4then find out the value of seca,tana,51sina+34cosa=?​

Answer 1

$\huge\mathfrak\red{Answer}$

$\sec \alpha + \tan \alpha = 4 \\ \\ \tan \alpha = 4 - \sec \alpha \\ {tan}^{2} \alpha = {(4 - sec \alpha )}^{2} \\ {tan}^{2} \alpha = 16 + {sec}^{2} \alpha - 8 \sec \alpha \\ {sec}^{2} \alpha - 1 = 16 + {sec}^{2} \alpha - 8 \sec \alpha \\ - 17 = - 8 \sec \alpha \\ \\ \sec \alpha = \frac{17}{8} ........(1) \\ \\ sec \alpha + tan \alpha = 4 \\ \frac{17}{8} + tan \alpha = 4 \\ \tan \alpha = 4 - \frac{17}{8} \\ \tan \alpha = \frac{15}{8} \\ \\ \tan \alpha = \frac{15}{8} .........(2) \\ \\ \sec \alpha = \frac{1}{ \cos \alpha } \\ \frac{1}{cos \alpha } = \frac{17}{8} \\ \cos \alpha = \frac{8}{17} .........(3) \\ \\ { \cos }^{2} \alpha + {sin}^{2} \alpha = 1 \\ {( \frac{8}{17}) }^{2} + {sin}^{2} \alpha = 1 \\ \frac{64}{289} + {sin}^{2} \alpha = 1 \\ \\ {sin}^{2} \alpha = 1 - \frac{64}{289} \\ {sin}^{2} \alpha = \frac{289 - 64}{289} \\ \\ {sin}^{2} \alpha = \frac{225}{64} \\ \\ \sin \alpha = \frac{15}{8} ........(4) \\ \\ 51 \sin \alpha + 34 \cos \alpha \\ \\ using \: (3) \: and \: (4) \\ 51( \frac{15}{8} ) + 34( \frac{8}{17} ) \\ \frac{51 \times 15}{8} + 16 \\ \frac{765 + 8 \times 16}{8} \\ \frac{893}{8}$

Answer 2

Answer:

u can use the trigonometry formula for this

hope before answer will help u lot