Question

Sec theta/ tan theta + cot theta= sin theta, how?

Answer 1

\[\large \frac{sec}{tan + cot}= sin\]

lets change everything to sin and cos

\[\Large \frac{\frac{1}{cos}}{\frac{sin}{cos} + \frac{cos}{sin}} = sin\]

simplify the denominator
\[\Large \frac{\frac{1}{cos}}{\frac{sin^2}{\cos\sin} + \frac{cos^2}{\cos\sin}} = sin\]

put it over the common denominator
\[\Large \frac{\frac{1}{cos}}{\frac{sin^2 + cos^2}{\cos\sin}} = sin\]

\[\large \frac{\cancel{\cos} \sin}{\cancel{cos}(sin^2 + cos^2)} = sin\]

\[\large \frac{sin}{sin^2 + cos^2} = sin\]

remember the identity \(\large sin^2(x) + cos^2(x) = 1\) so we have

\[\large \frac{sin}{1} = sin\]
\[\large \sin = \sin \checkmark\]

Answer 2

Thanks that helped

Answer 3

no problem!