Question

3tanA+4=0 then find 2cotA-5cosA+sinA

Answer 1

hello,
given that,
3tanA + 4=0
3tanA=-4
tanA=-4/3
or
cotA=3/-4
here, let Perpendicular be -4k and base be 3k
-Applying Pythagoras theorem to find hypotenuse...
$( - 4k)^{2} + (3k)^{2} = (h)^{2} \\ 16 {k}^{2} + 9 {k}^{2} = {h}^{2} \\ 25 {k}^{2} = {h}^{2} \\ 5k = h$
sinA= -4k/5k
=-4/5
cosA=3k/5k
=3/5
putting values in question=
$= 2 \times ( \frac{3}{ - 4} ) -5 \times \frac{3}{5} + ( \frac{ - 4}{5} ) \\ = - \frac{3}{2} - 3 - \frac{4}{5} \\ = - ( \frac{3}{2} + 3 + \frac{4}{5} )$
$= - (\frac{15 + 30 + 8}{10} ) \\ = - \frac{53}{10} \\ = - 5.3$