(Sec(theta)*sin(theta))/(tan(theta)+cot(theta))=sin^2(theta)
Really?
Just cancel the cosx , top and bottom
sec θ ∙ sin θ / ( tan θ + cot θ) =
( 1 / cos θ ) ∙ sin (θ) / ( sin θ / cos θ + cos θ / sin θ )n=
( 1 / cos θ ) ∙ sin θ / [ ( sin θ ∙ sin θ + cos θ ∙ cos θ ) / ( cosθ ∙ sin θ ) ] =
( 1 / cos θ ) ∙ sin θ / [ ( sin² θ + cos² θ ) / ( cosθ ∙ sin θ ) ] =
( 1 / cos θ ) ∙ sin θ / [ 1 / ( cosθ ∙ sin θ ) ] =
( 1 / cos θ ) ∙ sin θ ∙ 1 / [ 1 / ( cosθ ∙ sin θ ) ] =
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Remark:
1 / ( 1 / x ) = x
that's why
1 / [ 1 / ( cosθ ∙ sin θ ) ] = cosθ ∙ sin θ
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( 1 / cos θ ) ∙ sin θ ∙ cosθ ∙ sin θ =
( 1 / cos θ ) ∙ cosθ ∙ sin θ ∙ sin θ =
( cos θ / cos θ ) ∙ sin θ ∙ sin θ =
1 ∙ sin² θ = sin² θ
To explain how to prove the equation (sec(theta) * sin(theta))/(tan(theta) + cot(theta)) = sin^2(theta), we need to simplify the left-hand side (LHS) and the right-hand side (RHS) and then show that they are equal.
Let's start with the LHS:
(sec(theta) * sin(theta))/(tan(theta) + cot(theta))
Now, let's rewrite sec(theta), tan(theta), and cot(theta) in terms of sin(theta) and cos(theta):
sec(theta) = 1/cos(theta)
tan(theta) = sin(theta)/cos(theta)
cot(theta) = cos(theta)/sin(theta)
Substituting these values, we have:
(1/cos(theta) * sin(theta))/(sin(theta)/cos(theta) + cos(theta)/sin(theta))
Next, let's simplify the denominator by finding a common denominator:
(sin(theta)/cos(theta) + cos(theta)/sin(theta)) = (sin^2(theta) + cos^2(theta))/(cos(theta) * sin(theta)/cos(theta))
Since sin^2(theta) + cos^2(theta) = 1, we have:
(1)/(cos(theta) * sin(theta)/cos(theta))
Now, dividing by a fraction is the same as multiplying by its reciprocal, so we can rewrite the expression as:
1 * (cos(theta)/cos(theta))/(cos(theta) * sin(theta))
The cos(theta) terms in the numerator and denominator cancel out:
1 * 1/(cos(theta) * sin(theta))
So, the simplified LHS is:
1/(cos(theta) * sin(theta))
Now, let's simplify the RHS:
sin^2(theta)
Since sin^2(theta) is already simplified, we don't need to do any further calculations.
Comparing the simplified LHS and RHS, we can see that they are indeed equal:
1/(cos(theta) * sin(theta)) = sin^2(theta)
Hence, we have proved that (sec(theta) * sin(theta))/(tan(theta) + cot(theta)) = sin^2(theta).
I am going to use x for theta
(Secx*sinx)/(tanx+cotx)=sin^2 x
LS = (1/cosx * sinx)/(sinx/cosx + cosx/sinx)
= (sinx/cosx)/((sin^2 x + cos^2 x)/(sinxcosx))
= (sinx/cosx)(sinxcosx/1)
= can you see it happening?