Proving Equation: tan^2θ - sin^2θ = tan^2θsin^2θ

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To prove the equation tan²θ - sin²θ = tan²θsin²θ, start by recalling the definition of tangent, tan(θ) = sin(θ)/cos(θ), which leads to tan²(θ) = sin²(θ)/cos²(θ). By substituting this into the equation, you can manipulate one side using trigonometric identities. The goal is to simplify the expression until both sides of the equation match. This involves algebraic manipulation and applying fundamental trigonometric relationships.
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Homework Statement


Prove that tan^2θ - sin^2θ = tan^2θsin^2θ


Homework Equations


I'm not sure :S


The Attempt at a Solution


I have no idea
 
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What is the definition of tan?
 
tan = sin^2θ/cos^2θ

?
 
\tan (\theta) = \frac{\sin(\theta)}{\cos(\theta)}

so

\tan^2 (\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}

You will then need to manipulate one side until it equals the other using basic trigonometric identities.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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