Question

8sin square 45 -2 tan square 60 +3 cot square 30 -2 cos square 45 =​

Answer 1

Answer:

0 is the answer of this question

Answer 2

$\sf \implies8 \: {sin}^{2} 45 \degree - 2{tan}^{2} 60\degree + 3{cot}^{2} 30\degree - 2 {cos}^{2} 45\degree$

We know that :-

• Sin 45° = 1/√2

• tan 60° = √3

• cot 30 = √3

• cos 45 = 1/√2

$\sf \implies8 \: {( \dfrac{1}{ \sqrt{2} }) }^{2} - 2{ (\sqrt{3} )}^{2} + 3{ (\sqrt{3} )}^{2} - 2 { (\dfrac{1}{ \sqrt{2} }) }^{2}$

$\sf \implies8 \: {( \dfrac{1}{ 2 }) } - (2 \times 3) + (3 \times 3) - 2 { (\dfrac{1}{ {2} }) }$

$\sf \implies4 -6+ 9- 1$

$\sf \implies6$

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$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

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Learn more:-

1. Cosθ = base / hypotenuse

2. cossecθ = 1/ sinθ

3. sec θ = 1/cosθ

4. Cotθ = 1/ tanθ

5. Sin²θ+ Cos²θ= 1

6. Sec²θ - tan²θ = 1

7. cosec ²θ - cot²θ = 1

8. sin(90°−θ) = cos θ

9. cos(90°−θ) = sin θ

10. tan(90°−θ) = cot θ

11. cot(90°−θ) = tan θ

12. sec(90°−θ) = cosec θ

13. cosec(90°−θ) = sec θ

14. Sin2θ = 2 sinθ cosθ

15. cos2θ = Cos²θ- Sin²θ