Question

if cosA =3/5, find 9cot2A-1

Answer 1

Heya !!!


CosA = 3/5 = B/H




B = 3 and H = 5


By Pythagoras theroem ,



(H)² = (B)²+(P)²

(P)² = (H)²-(B)²

(P)² = (5)²-(3)² = 25-9


P = ✓16 = 4


Therefore,


CotA = B/P = 3/4




9 Cot²A - 1 = 9 × (3/4)² -1



=> 9 × 9/16 - 1



=> 81/16 - 1

=> 81-16/16


=> 65/16



★ HOPE IT WILL HELP YOU ★

Answer 2

Hiii. ....friend

Here is ur answer,

CosA => 3/5.

CosA= adj side/Hypotenuse.

Adj side => 3.

Hypotenuse => 5.

Opp side = ?

$= > \: {hypo}^{2} = {base}^{2} + {height}^{2}$

$= > \: h \: = \sqrt{ {5}^{2} - {3}^{2} }$

$= > \: \sqrt{25 - 9} = \sqrt{16}$

$= > \: 4$

CotA=> Adj side/ Opp side.

=> 3/ 4.


=> 9Cot2A-1

$= > \: \frac{9 {cot}^{2}a - 1 }{2 \cot(a) } - 1$

$= > \: \frac{9 { (\frac{3}{4} })^{2} - 1}{2( \frac{3}{4} )} - 1$

$= > \: \frac{ \frac{81}{16} - 1}{ \frac{3}{2} } - 1$

$= > \: \frac{65}{16} \times \frac{2}{3} - 1$

$= > \: \frac{65}{24} - 1$

$= > \: \frac{41}{24}$



:-)Hope it helps u.