Question

if tanA = -4/3 and A doesn't lie in 4th quadrant then show that 5sinA + 10cosA + 9secA + 16 cosecA + 4cotA = 0​

Answer 1

Answer:

i don't know about it sorry

Answer 2

Answer:

★Given:

1. 1st case:
2. ${ \tan(a) }^{2} = \frac{16}{9} \\ { \cot(a) }^{2} = \frac{9}{16} \\ 1 + { \cot(a) }^{2} = \frac{9}{16} + 1 \\ { \csc(a) }^{2} = \frac{25}{16} \\ \csc(a) = \frac{5}{4} \\ \sin(a) = \frac{4}{5}$

$\tan a = - \frac{4}{3} \\ { \tan(a) }^{2} = \frac{16}{9} \\ 1 + { \tan(a) }^{2} = \frac{16}{9 } + 1 \\ { \sec(a) }^{2} = \frac{25}{9} \\ \sec(a) = \frac{5}{3} \\ \cos(a) = \frac{3}{5}$

Proving area:

N.B if tanA doesn't lie in 4 th quadrant the if it is minus so this is lie in 2nd quadrant as the observation.

a doesn't lie in 4 th quadrant.

L.H.S

$= 5 \sin \: a + 10 \cos \: a + 9sec \: a + 16 \: \csc \: a + 4 \cot \: a$

$= 5 \times \frac{4}{5} + 10 \times - \frac{3}{5} + 9 \times - \frac{5}{3} + 16 \times \frac{5}{4} + 4 \times - \frac{3}{4}$

$= 4 - 6 - 15 + 20 - 3 \\ = 24 - 24 \\ = 0$

=R.H.S (Proved)

Hope you understand.