Explanation:
We need to verify:
[tex]cos\theta-sec\theta= -sin\theta tan\theta \\ \\ \\ \text{We know:} \\ \\ sec \theta=\frac{1}{cos \theta} \\ \\ \\ Then: \\ \\ cos\theta-\frac{1}{cos \theta}=\frac{cos^2\theta-1}{cos\theta} \\ \\ \\ Remember \ that: \\ \\ sin^2\theta +cos^2\theta=1 \therefore sin^2\theta=1-cos^2\theta \therefore -sin^2\theta=cos^2\theta -1 \\ \\ \\ Therefore: \\ \\ \frac{cos^2\theta-1}{cos\theta}=-\frac{sin^2\theta}{cos\theta} \\ \\ \\ But: \\ \\ sin^2\theta=sin\theta sin\theta[/tex]
[tex]Then: \\ \\ -\frac{sin^2\theta}{cos\theta}=-\frac{sin\theta sin\theta}{cos\theta} \\ \\ \\ And: \\ \\ \frac{sin\theta}{cos\theta}=tan\theta \\ \\ \\ Finally: \\ \\ -\frac{sin\theta sin\theta}{cos\theta}=-sin\theta tan\theta[/tex]
Conclusion:
[tex]cos\theta-sec\theta= -sin\theta tan\theta[/tex]
