Question

1, tan 45° + 2 tan² 60° = 
2, sin² 30° = 1 = 
3, tan² 60° + 4 cos ² 45° + 3 sec² 30° + 5 cos² 90° =   ​

Answer 1

Answer:

1 ) tan60°=

$\sqrt{3}$

tan45∘ =1

$tan^2 60∘ +tan^2 45∘ =( \sqrt{3} )^2+(1)^2 =3+1=4$

Hence the value is 4.

2)

$sin^2 x+sin^2 30=1 \\ </p><p></p><p>sin ^{2} x = 1 - sin^{2} 30sin^2 x=1−sin^2 30 \\ </p><p></p><p>sin ^{2} x= 1 - 1 \div 4sin^2 x=1−1÷4 \\ </p><p></p><p>sin ^{2} x= ( \ \sqrt{3} ) \div 2sin^2</p><p> x=( \sqrt{3} )÷2 \\ </p><p></p><p>sin ^{2} x = 60 ^{o}sin^2</p><p> x=60°$

3)

$by \: substituting \: the \: value \: we \: get \\ \frac{ \sqrt{3} ^{2} + 4( \frac{1}{ \sqrt{2} })^{2} + 3 (\frac{2}{ \sqrt{3} } )^{2} + 5}{2 + 2 - \sqrt{3} ^{2} } \\ \\ \frac{3 + 2 + 4}{2 + 2 - 3} \\ \\ \frac{9}{1} \\ = 9$

Explanation:

hope it helps :)