6. Prove that: \( (\sin \theta+\cos \theta)^{2}=1+2 \sin \theta \cos \theta \) 8. Prove that: \( \frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}=\frac{1-\cos \theta}{1+\cos \theta} \) 10. Prove that: \( \frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=\frac{4 \operatorname{con} x}{\cos x} \) 11. Prove that: \( \frac{1+2 \sin \theta . \cos \theta}{\sin \theta+\cos \theta}=\sin \theta+\cos \theta \) (consider number 6\( ) \)
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To prove the identity \( (\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta \), let's start by expanding the left side. We know that \( (a + b)^2 = a^2 + 2ab + b^2 \). Here, \( a = \sin \theta \) and \( b = \cos \theta \), so we get: \[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta \] Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute this into our equation: \[ = 1 + 2 \sin \theta \cos \theta \] Thus, we have successfully proven that \( (\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta \). Now onto some potential pitfalls to avoid while proving identities like these! Always remember to utilize known identities such as the Pythagorean identity or double angle formulas, as they can simplify your work and lead you to the solution more quickly. Also, be mindful of squaring both sides: adding extra conditions can sometimes lead to errors in your proofs, so check for extraneous solutions when rearranging terms!
