Prove that: $$ \frac{(\sin heta+\cos heta)^{2}}{\cos heta}-\frac{1}{\cos heta}=2 \cdot \sin heta $$

Question

Prove that: $$ \frac{(\sin 	heta+\cos 	heta)^{2}}{\cos 	heta}-\frac{1}{\cos 	heta}=2 \cdot \sin 	heta $$

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1. Rewrite the expression:
   $$
   \frac{(\sin \theta+\cos \theta)^{2}}{\cos \theta} - \frac{1}{\cos \theta}.
   $$

2. Combine the terms over the common denominator:
   $$
   \frac{(\sin \theta+\cos \theta)^{2} - 1}{\cos \theta}.
   $$

3. Expand the squared term in the numerator:
   $$
   (\sin \theta+\cos \theta)^{2} = \sin^{2}\theta + 2\sin \theta\cos \theta + \cos^{2}\theta.
   $$

4. Utilize the Pythagorean identity $$ \sin^{2}\theta + \cos^{2}\theta = 1 $$:
   $$
   \sin^{2}\theta + 2\sin \theta\cos \theta + \cos^{2}\theta = 1 + 2\sin \theta\cos \theta.
   $$

5. Substitute this back into the expression:
   $$
   \frac{(1 + 2\sin \theta\cos \theta) - 1}{\cos \theta} = \frac{2\sin \theta\cos \theta}{\cos \theta}.
   $$

6. Cancel $$ \cos \theta $$ in the numerator and denominator (assuming $$ \cos \theta \neq 0 $$):
   $$
   2\sin \theta.
   $$

Thus, the given identity is proven:
$$
\frac{(\sin \theta+\cos \theta)^{2}}{\cos \theta} - \frac{1}{\cos \theta} = 2\sin \theta.
$$
$$ \frac{(\sin \theta+\cos \theta)^{2}}{\cos \theta} - \frac{1}{\cos \theta} = 2\sin \theta $$

Prove that: \( \frac{(\sin \theta+\cos \theta)^{2}}{\cos \theta}-\frac{1}{\cos \theta}=2 \cdot \sin \theta \)

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