which pythagorean identity is correct?\n○ $sin^{2}(\theta)+1 = cos^{2}(\theta)$\n○ $\tan^{2}(\theta)+1=sec^{2…

Explanation:

Step1: Recall Pythagorean identities

The fundamental Pythagorean identity in trigonometry is $\sin^{2}(\theta)+\cos^{2}(\theta) = 1$. Dividing this identity by $\cos^{2}(\theta)$ gives $\frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}+\frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}=\frac{1}{\cos^{2}(\theta)}$. Since $\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta)$ and $\frac{1}{\cos(\theta)}=\sec(\theta)$, we get $\tan^{2}(\theta)+ 1=\sec^{2}(\theta)$. Dividing $\sin^{2}(\theta)+\cos^{2}(\theta) = 1$ by $\sin^{2}(\theta)$ gives $1+\cot^{2}(\theta)=\csc^{2}(\theta)$.

Step2: Analyze each option

• Option 1: $\sin^{2}(\theta)+1=\cos^{2}(\theta)$ is incorrect as $\sin^{2}(\theta)+\cos^{2}(\theta)=1$.
• Option 2: $\tan^{2}(\theta)+1=\sec^{2}(\theta)$ is a correct Pythagorean - related identity as derived above.
• Option 3: The correct identity is $1 + \cot^{2}(\theta)=\csc^{2}(\theta)$, not $1-\cot^{2}(\theta)=\csc^{2}(\theta)$.
• Option 4: From $\sin^{2}(\theta)+\cos^{2}(\theta)=1$, we have $\sin^{2}(\theta)=1 - \cos^{2}(\theta)$, and $\tan^{2}(\theta)=\frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}$, so $1-\cos^{2}(\theta)\neq\tan^{2}(\theta)$.

Answer:

$\tan^{2}(\theta)+1=\sec^{2}(\theta)$