Question

1) wiedzac ze tgα=4/5 i α jest kątem ostrym, oblicz sinα+cosα

2) wiedząc ze ctgα=5/6 i α jest kątem ostrym, oblicz sinα+cosα

3) sprawdź czy rownosc sinα+cosα*ctgα=1/sinα jest prawdziwa

Answer 1

1)

tg a = 4/5

tg a = y/x , zatem   y = 4  i  x= 5

r^2 = x^2 + y^2 = 5^2 + 4^2 = 25 + 16 = 41

r = p (41)

sin a = y/r = 4/p(41)

cos a = x/r = 5/p(41)

zatem

sin a + cos a = 4/p(41) + 5/p(41) = 9/ p(41) = ( 9 p(41))/ 41

======================================================

2)

ctg a = 5/6

ctg a = x/y;  zatem   x = 5  i  y = 6

r^2 = x^2 + y^2  = 5^2 + 6^2 = 25 + 36 = 61

r = p(61)

sin a = y/r = 6/ p(61)

cos a = x/r = 5/ p(61)

zatem

sin a + cos a = 6/ p(61) + 5/p(61) = 11/ p(61) = ( 11 p(61))/61

=======================================================

3)

L = sin a + cos a * ctg a = sin a + cos a *[ cos a /sin a] =

= sin a + cos ^2 a/ sin a = sin^2 a / sin a + cos ^2 a/ sin a =

= [ sin ^2 a + cos^ 2 a ]/sin a = 1 / sin a = P

=========================================