Question

if sin ∅ A = 3/5 and cos B = 12/13 find sec² A and tan A + tan b​

Answer 1

given=>

sin ∅ A = 3/5 and cos B = 12/13 f

let's A=a

B=c

to find=>

we know that

$= > \sin( a ) = \frac{3}{5} = \frac{perpendicular}{hypotenius} \\ \\ = > \cos(c) = \frac{12}{13} = \frac{base}{hypoteniuse} \\ \\ we \: know \: that \: paythagorus \: rules \\ \\ = > ( a) = h = \sqrt{ {b {}^{2} + p {}^{2} } } \\ \\ = > b = \sqrt{h {}^{2} - p {}^{2} } \\ \\ = > b = \sqrt{5 {}^{2} - 3 {}^{2} } \\ \\ = > b = \sqrt{16} \\ \\ = > b = 4 \\ \\ \cos(a) = \frac{b}{h} = \frac{4}{5} \\ \\ now \: in \: c \\ \\ = > p = \sqrt{13 {}^{2} - 1 2{}^{2} } \\ \\ = > p = \sqrt{25} \\ \\p = 5 \\ \\ \sec(c) = \frac{h}{b} \\ \\ \sec(c) = \frac{13}{12} \\ \\ = \sec {}^{2} (c) = ( \frac{13}{12} ) {}^{2} = \frac{169}{144} \\ \\ \\ other \: rules \: we \: know \: thar \\ \\ \cos(c) = \frac{1}{ \sec(c) } \\ \\ when \\ \cos(c) = \frac{12}{13} \\ \\ to \sec(c) \frac{13}{12}$

tan a=3/4, tan(c) =5/12

tan(a) +tan(c)

3/4+5/12

7/6

________________________

I hope it helps you