How do tractor beams work?
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✳️ copy paste answer ignored...
Answers
Explanation:
\LARGE{\bf{\underline{\underline\color{blue}{GIVEN:-}}}}
GIVEN:−
\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}∙
(1+sinA+cosA)
2
(1+sinA−cosA)
2
=
1+cos
1−cos
\LARGE{\bf{\underline{\underline\color{blue}{SOLUTION:-}}}}
SOLUTION:−
LHS:
\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→→
(1+sinA+cosA)
2
(1+sinA−cosA)
2
→
Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.
\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→→
(cos
2
+2sincos+sin
2
+2cos+2sin+1)
(cos
2
−2sincos+sin
2
−2cos+2sin+1)
→
Rearrange the terms.
\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→→
(cos
2
+sin
2
+2sincos+2cos+2sin+1)
(cos
2
+sin
2
−2sincos−2cos+2sin+1)
→
We know that cos²A+sin²A=1.
\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→→
2sin+1
1−2sincos−2cos
→
Now here, take -2cos common from the numerator and +2cos common from the denominator.
\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→→
2sin+1
1−2cos(sin+2)
→
Now, rearrange the terms, add 1 and 1 and take 2 common.
\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→→
sin+1
1+1+2sin−2cos
→
\to\sf\dfrac{2+2sin-2cos}{sin+1}→→
sin+1
2+2sin−2cos
→
Take 2 common.
\to \sf \dfrac{ 2(1+sin) -2cos(sin+1) }{ 2(1+sin) + 2cos(sin +1 ) }→→
2(1+sin)+2cos(sin+1)
2(1+sin)−2cos(sin+1)
→
\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→ < /p > < p >→
1+cosA
1−cosA
→</p><p>
LHS=RHS.
HENCE PROVED!
FUNDAMENTAL TRIGONOMETRIC RATIOS:
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}\end{gathered}\end{gathered} \end{gathered}
sin
2
θ+cos
2
θ=1
1+cot
2
θ=cosec
2
θ
1+tan
2
θ=sec
2
θ
T-RATIOS:
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\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}}}\end{gathered}\end{gathered} \end{gathered}
∠A
sinA
cosA
tanA
cosecA
secA
cotA
0
∘
0
1
0
NotDefined
1
NotDefined
30
∘
2
1
2
3
3
1
2
3
2
3
45
∘
2
1
2
1
1
2
2
1
60
∘
2
3
2
1
3
3
2
2
3
1
90
∘
1
0
NotDefined
1
NotDefined
0
Answer:
This practical tractor beam uses ultrasonic waves to levitate small spheres at short distances. ... Optical tractor beams (using lasers) have been around for a while, but they don't exert much force for the energy you expend to create them.
Explanation:
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