Prove that: $$ \frac{\sin heta-\cos heta \sin heta}{\cos heta-\left(1-\sin ^{2} heta ight)}= an heta $$ Determine the general solution of $$ 2 \sin x \cos x-\cos ^{2} x $$

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Prove that: $$ \frac{\sin 	heta-\cos 	heta \sin 	heta}{\cos 	heta-\left(1-\sin ^{2} 	hetaight)}=	an 	heta $$ Determine the general solution of $$ 2 \sin x \cos x-\cos ^{2} x $$

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We will solve the two parts separately.

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Part 1. Prove that 
  (sin θ – cos θ · sin θ) / (cos θ – (1 – sin²θ)) = tan θ

Step 1. Factor the numerator.  
Notice that in the numerator sin θ – cos θ · sin θ, you can factor sin θ:  
  sin θ – cos θ · sin θ = sin θ (1 – cos θ).

Step 2. Simplify the denominator.  
The denominator is cos θ – (1 – sin²θ). Begin by expanding:
  cos θ – 1 + sin²θ.
Now recall the Pythagorean identity sin²θ = 1 – cos²θ. Substituting this we get:
  cos θ – 1 + (1 – cos²θ) = cos θ – cos²θ = cos θ (1 – cos θ).

Step 3. Combine the results.  
So the original expression becomes:
  [sin θ (1 – cos θ)] / [cos θ (1 – cos θ)].

Provided that (1 – cos θ) ≠ 0 (i.e. cos θ ≠ 1) and cos θ ≠ 0, you can cancel the common factor (1 – cos θ) to get:
  sin θ / cos θ = tan θ.

Thus, we have proved that
  (sin θ – cos θ sin θ) / (cos θ – (1 – sin²θ)) = tan θ.

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Part 2. Determine the general solution of 2 sin x cos x – cos² x

Often, when asked to “determine the general solution” of an expression in trigonometry, it is understood that we set the expression equal to zero and solve. Thus, we start with:
  2 sin x cos x – cos² x = 0.

Step 1. Factor the expression.  
Notice that cos x is a common factor:
  cos x (2 sin x – cos x) = 0.

This product equals zero when either factor equals zero.

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Case 1: cos x = 0

The cosine function equals zero at:
  x = π/2 + πk, where k is any integer.

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Case 2: 2 sin x – cos x = 0

Solve for sin x:
  2 sin x = cos x  
Divide both sides by cos x (assuming cos x ≠ 0):
  2 tan x = 1  
That is:
  tan x = 1/2.

The general solution for tan x = 1/2 is:
  x = arctan(1/2) + πk, where k is any integer.

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Thus, the general solution of 2 sin x cos x – cos² x = 0 is:
  x = π/2 + πk  or  x = arctan(1/2) + πk, where k ∈ ℤ.

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Summary of Answers

1. We have shown that:
  (sin θ – cos θ sin θ) / (cos θ – (1 – sin²θ)) = tan θ, (with appropriate restrictions on θ).

2. The general solution of 2 sin x cos x – cos² x = 0 is:
  x = π/2 + πk  or  x = arctan(1/2) + πk, for any integer k.
To prove that $$ \frac{\sin 	heta - \cos 	heta \sin 	heta}{\cos 	heta - (1 - \sin^2 	heta)} = 	an 	heta $$, factor the numerator to get $$ \sin 	heta (1 - \cos 	heta) $$ and simplify the denominator to $$ \cos 	heta (1 - \cos 	heta) $$. Canceling the common factor $$ (1 - \cos 	heta) $$ (assuming $$ \cos 	heta 
eq 1 $$ and $$ \cos 	heta 
eq 0 $$), we obtain $$ \frac{\sin 	heta}{\cos 	heta} = 	an 	heta $$. For the equation $$ 2 \sin x \cos x - \cos^2 x = 0 $$, factor out $$ \cos x $$ to get $$ \cos x (2 \sin x - \cos x) = 0 $$. This gives two cases: 1. $$ \cos x = 0 $$, which has solutions $$ x = \frac{\pi}{2} + \pi k $$ for any integer $$ k $$. 2. $$ 2 \sin x - \cos x = 0 $$, leading to $$ 	an x = \frac{1}{2} $$, with solutions $$ x = \arctan\left(\frac{1}{2}ight) + \pi k $$ for any integer $$ k $$. Thus, the general solutions are $$ x = \frac{\pi}{2} + \pi k $$ or $$ x = \arctan\left(\frac{1}{2}ight) + \pi k $$, where $$ k $$ is any integer.

Prove that: \( \frac{\sin \theta-\cos \theta \sin \theta}{\cos \theta-\left(1-\sin ^{2} \theta\right)}=\tan \theta \) Determine the general solution of \( 2 \sin x \cos x-\cos ^{2} x \)

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