Question

verify using sin and cos

(tanθ+cotθ)tanθ=sec^2θ

Answer 1

convert it into terms of sin and cos

Answer 2

You could use \[
\tan(\theta)\cot(\theta) = 1
\]

Answer 3

You take advantage of (tanx)^2 = (secx)^2 - 1. This is a derivation of a pythagorean identity of trig functions. As @wio said, tant * cott = 1. Thus the sum is (secx)^2 + 1 - 1 = (secx)^2

Answer 4

ok, I understand. But is there a way to use only sin and cos?

Answer 5

\[\tan \theta = \frac{ \sin \theta }{\cos \theta }, \cot \theta = \frac{\cos \theta}{\sin \theta}\]

so

\[(\tan \theta + \cot \theta)\tan \theta = \frac{ \sin ^{2}\theta }{ \cos ^{2} \theta} + 1\] which equals \[\frac{ \sin^{2}\theta + \cos^{2} \theta }{\cos^{2}\theta }\] by the Pythagoean identity, we know that the numerator equals 1, thus we get
\[\frac{ 1 }{ \cos^{2}\theta } = \sec^{2} \theta\] by definition

Answer 6

ok, I got it. thanks for both of your help!