Question

NEED IT ASAP
if tanA/root3 = 1, the value of 2sin^2A - 3cos^2A

Answer 1

$\frac{ \tan(A) }{ \sqrt{3} } = 1$

$\implies \tan(A) = \sqrt{3}$

$\implies A = 60\degree$

$2 \sin^{2} ( A ) - 3 \cos ^{2} (A)$

$\implies 2 \sin^{2} ( 60 ) - 3 \cos ^{2} (60)$

$\implies 2 {( \frac{ \sqrt{3} }{2} )}^{2} - 3 {( \frac{1}{2} )}^{2}$

$\implies 2 ( \frac{ 3}{4} ) - 3 ( \frac{1}{4} )$

$\implies \frac{ 6}{4} - \frac{3}{4}$

$\orange{ \therefore \frac{ 3}{4} }$

Answer 2

Answer:

$2 { \sin }^{2} A - 3 { \cos }^{2} A = \frac{3}{4}$

Step-by-step explanation:

Given that:-

$\frac{ \tan(A) }{ \sqrt{3} } = 1$

$\tan(A) = \sqrt{3}$

$A = 60°$

So,

$2 { \sin }^{2} A - 3 { \cos }^{2} A \\ = 2 { (\sin60°)}^{2} - 3 {( \cos60°) }^{2}$

$= 2 { (\frac{ \sqrt{3} }{2} )}^{2} - 3 { (\frac{1}{2}) }^{2}$

$= (2 \times \frac{3}{4} ) - (3 \times \frac{1}{4} )$

$= \frac{3}{2} - \frac{3}{4}$

$= \frac{6 - 3}{4} = \frac{3}{4}$