Question

3. Evaluate the following:
4/3 tan2 30° + sín2 60° - 3 cos2 60° +
tan2 60°3/4 - 2 tan2 45°
​

Answer 1

Just remember some values

tan30 = 1/√3

tan 60 = tan(90-30)= cot 30 = √3

cos60 = 1/2

tan 45 = 1

So

4/3 tan2 30 + sin2 60 - 3cos2 60 + tan2 60 3/4. - 2 tan2 45

= 4/3 × 1/3 + sin2 60 + cos2 60 - 4 cos2 60 + 9/4 - 2

= 4/9. + 1 - 4/4 + 9/4 -2

= 4/9 +1-1 +9/4 -2

= 16 + 81 - 72)/36

= 25/36

Answer 2

Answer:

Required value of the given expression is $\frac{25}{ 36}$

Step-by-step explanation:

Given,

$\frac{4}{3} {tan}^{2} {30}^{o} + {sin}^{2} {60}^{o} - 3 {cos}^{2} {60}^{o} + {tan}^{2} {60}^{o} \times \frac{3}{4} - 2 {tan}^{2} {45}^{o}$

We know,

$\tan( {30}^{o} ) = \frac{1}{ \sqrt{3} } \\ \tan( {60}^{o} ) = \sqrt{3} \\ \tan( {45}^{o} ) = 1 \\ \sin( {60}^{o} ) = \frac{ \sqrt{3} }{2} \\ \cos( {60}^{o} ) = \frac{1}{2}$

Now,

$\frac{4}{3} {tan}^{2} {30}^{o} + {sin}^{2} {60}^{o} - 3 {cos}^{2} {60}^{o} + {tan}^{2} {60}^{o} \times \frac{3}{4} - 2 {tan}^{2} {45}^{o}$

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

$= \frac{4}{3} \times \frac{1}{3} + \frac{3}{4} - \frac{3}{4} + 3 \times \frac{3}{4} - 2 \\ = \frac{4}{9} + \frac{9}{4} - 2 \\ = \frac{16 + 81 - 72}{36} \\ = \frac{25}{36}$

This is a problem of Trigonometry.

Some important Trigonometry formulas,

$sin(x) = \cos(\frac{\pi}{2} - x) \\ \tan(x) = \cot(\frac{\pi}{2} - x) \\ \sec(x) = \csc(\frac{\pi}{2} - x) \\ \cos(x) = \sin(\frac{\pi}{2} - x) \\ \cot(x) = \tan(\frac{\pi}{2} - x) \\ \csc(x) = \sec(\frac{\pi}{2} - x)$

know more about Trigonometry,

https://brainly.in/question/8632966

https://brainly.in/question/11371684

#SPJ2