Question

secA+tanA=tan(π/4+A/2) ​

Answer 1

Step-by-step explanation:

$\tt secA + tanA \\ \\ \tt \longrightarrow \frac{1}{cosA} + \frac{sinA}{cosA} \\ \\ \tt \longrightarrow \frac{1+sinA}{cosA} \\ \\ \tt \longrightarrow \frac{{sin}^{2}\frac{A}{2} + {cos}^{2}\frac{A}{2} + 2sin\frac{A}{2}cos\frac{A}{2}}{{cos}^{2}\frac{A}{2} - {sin}^{2}\frac{A}{2}} \\ \\ \tt \longrightarrow \frac{{(sin\frac{A}{2} + cos\frac{A}{2})}^{2}}{(sin\frac{A}{2} + cos\frac{A}{2})( cos\frac{A}{2} - sin\frac{A}{2})} \\ \\ \tt \longrightarrow \frac{sin\frac{A}{2} + cos\frac{A}{2}}{cos\frac{A}{2} - sin\frac{A}{2}} \\ \\ \tt Dividing\:by\:cos\frac{A}{2} \: \\ \\ \tt \longrightarrow \frac{1+tan\frac{A}{2}}{1-tan\frac{A}{2}} \\ \\ \tt \longrightarrow \frac{tan\frac{\pi}{4} + tan\frac{A}{2}}{1-tan\frac{A}{2}tan\frac{\pi}{4}} \\ \\ \tt \longrightarrow tan(\frac{\pi}{4}+\frac{A}{2}) \\ \\ \tt \longrightarrow R.H.S.$

Answer 2

secA + tanA

${sin}^{2} \frac{A}{2} + {cos}^{2} \frac{A}{2} + 2sin \frac{A}{2 }cos \frac{A}{2} {cos}^{2} \frac{A}{2} - {sin}^{2} \frac{A}{2}$

$= > \frac{1 + sinA}{cosA}$

$= > \frac{1}{cosA} + \frac{sinA}{cosA}$