Question

Given that tan Ѳ = sin Ѳ/ cos Ѳ , show that
1+ cot^2 Ѳ = csc^2Ѳ is a true trigonometric identity

Answer 1

\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \implies \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\), so \(\cot^2(\theta) = \frac{\cos^2(\theta)}{\sin^2(\theta)}\),

Adding one:

\(\cot^2(\theta) + 1 = \frac{\cos^2(\theta)}{\sin^2(\theta)} + \frac{\sin^2(\theta)}{\sin^2(\theta)}\)

\(=\frac{\cos^2(\theta) + \sin^2(\theta)}{\sin^2(\theta)}\)

\(=\frac{1}{\sin^2(\theta)}\)

\(= \csc^2(\theta)\)

Answer 2

I get it completely. Thank you very much.

Answer 3

No problem :).