Answer:
The simplified expression is:
\(\dfrac{1}{\cot^2 \theta}+\sec \theta\cos \theta=\sec^2 \theta\)
Step-by-step explanation:
We are asked to simplify the expression:
one divided by cotangent of theta to the second power+ sec θ cos θ
i.e. mathematically it is written as:
\(=\dfrac{1}{\cot^2 \theta}+\sec \theta\cos \theta\)
We know that:
\(\tan \theta=\dfrac{1}{\cot \theta}\\\\i.e.\\\\(\tan \theta)^2=(\dfrac{1}{\cot \theta})^2\\\\i.e.\\\\\tan^2 \theta=\dfrac{1}{\cot^2 \theta}\)
Hence, we can write this expression as:
\(\dfrac{1}{\cot^2 \theta}+\sec \theta\cos \theta=\tan^2 \theta+\sec \theta\cos \theta\)
Also, we know that:
\(\sec \theta=\dfrac{1}{\cos \theta}\)
Hence, we have:
\(\sec \theta\cos \theta=\dfrac{\cos \theta}{\cos \theta}\\\\\\i.e.\\\\\\\sec \theta\cos \theta=1\)
Hence, we get:
\(=\dfrac{1}{\cot^2 \theta}+\sec \theta\cos \theta=\tan^2 \theta+1\)
Also, we know that:
\(\sec^2 \theta-\tan^2 \theta=1\\\\i.e.\\\\\sec^2 \theta=1+\tan^2 \theta\)
Hence, we get the simplified expression as:
\(\dfrac{1}{\cot^2 \theta}+\sec \theta\cos \theta=\sec^2 \theta\)
