Question

Prove that $(co\sec A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1$.

Answer 1

LHS$=(co\sec A-\sin A)(\sec A-\cos A)(\tan A+\cot A)$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)(\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A})$

Using $\sec A=\frac{1}{\cos A}$ and $\csc A=\frac{1}{\sin A}$ and $[\frac{\sin \theta }{\cos \theta }=\tan \theta ]$

$=\frac{1-{\sin }^{2}A}{\sin A}\times \frac{1-{\cos }^{2}A}{\cos A}\times \frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}$

$=\frac{{\cos }^{2}A}{\sin A}\times \frac{{\sin }^{2}A}{\cos A}\times \frac{1}{\sin A\cos A}$ ............Using $[sin²\theta +cos²\theta =1]$

$=1$

Hence proved