Question

please solve my question
find 2 sin square 30 degree minus 3 cos square 45 degree + tan square 60 degree

Answer 1

sin 30°=1/2=>sin^2 30°=1/4
cos 45°=1/root 2=>cos^2 45°=1/2
tan 60°=root 3=>tan ^2 60°=3
So, 2 sin ^2 30°-3cos^2 45° +tan^2 60°
=>2(1/4)-3(1/2)+3
=>1/2-3/2+3
=>(1-3+6)/2
=>4/2
=>2

Answer 2

Answer:

The value of $2sin^2{30^{\circ}-3cos^2{45^{\circ}}+tan^2{60^{\circ}}$ is 2

Step-by-step explanation:

we have to find the value of

$2sin^2{30^{\circ}-3cos^2{45^{\circ}}+tan^2{60^{\circ}}$

$sin30^{\circ}=\frac{1}{2}\\cos45^{\circ}=\frac{1}{\sqrt2}\\tan60^{\circ}=\sqrt3$

Hence, $2sin^2{30^{\circ}-3cos^2{45^{\circ}}+tan^2{60^{\circ}}$

   = $2(\frac{1}{2})^2-3(\frac{1}{\sqrt2})^2+(\sqrt3)^2$

   = $\frac{1}{2}-\frac{3}{2}+3$

   =-1+3=2

Therefore, the value of $2sin^2{30^{\circ}-3cos^2{45^{\circ}}+tan^2{60^{\circ}}$ is 2