Question

(2tanθ-3cosθ) Expand

Answer 1

`\begin{gathered} 2\tan \theta-3\cos \theta=2(\sin\theta)/(\cos\theta)-3\cos \theta \\ 2(\sin\theta)/(\cos\theta)-3\cos \theta=2(\sin\theta)/(\cos\theta)-3(\cos ^2\theta)/(\cos \theta) \\ 2(\sin\theta)/(\cos\theta)-3(\cos^2\theta)/(\cos\theta)=(2\sin \theta-3\cos ^2\theta)/(\cos \theta) \\ (2\sin\theta-3\cos^2\theta)/(\cos\theta)=\frac{2\sin \theta-3(1-\sin ^2\theta_{})}{\cos \theta} \\ \frac{2\sin\theta-3(1-\sin^2\theta_{})}{\cos\theta}=(3\sin ^2\theta+2\sin \theta-3)/(\cos \theta) \end{gathered}`