Simplify: \(\frac{1+\tan^{2}(x)}{1+\cot^{2}(x)} \times \frac{\sec(x)-\tan(x)}{\sec(x)+\tan(x)}\)
\(\frac{\tan^{2}(x)(1-\sin(x))}{1+\sin(x)}\)
We simplify \(\frac{1+\tan^{2}(x)}{1+\cot^{2}(x)} \times \frac{\sec(x)-\tan(x)}{\sec(x)+\tan(x)}\) using standard trigonometric identities.
Recall the Pythagorean identities \(1+\tan^{2}(x)=\sec^{2}(x)\) and \(1+\cot^{2}(x)=\csc^{2}(x)\).
So the first fraction becomes \(\frac{\sec^{2}(x)}{\csc^{2}(x)} = \frac{1/\cos^{2}(x)}{1/\sin^{2}(x)} = \frac{\sin^{2}(x)}{\cos^{2}(x)} = \tan^{2}(x)\).
Next write \(\sec(x)-\tan(x) = \frac{1-\sin(x)}{\cos(x)}\) and \(\sec(x)+\tan(x) = \frac{1+\sin(x)}{\cos(x)}\).
Their ratio is \(\frac{1-\sin(x)}{\cos(x)} \div \frac{1+\sin(x)}{\cos(x)} = \frac{1-\sin(x)}{1+\sin(x)}\), since the \(\cos(x)\) terms cancel.
Multiplying the two simplified parts: \(\tan^{2}(x) \times \frac{1-\sin(x)}{1+\sin(x)} = \frac{\tan^{2}(x)(1-\sin(x))}{1+\sin(x)}\).
The key concept is converting sec, tan and cot into sine and cosine and applying the Pythagorean identities.
Hence the result is \(\frac{\tan^{2}(x)(1-\sin(x))}{1+\sin(x)}\).
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