Question

Prove that $\left(\cos ecA-\sin A\right)\left(\sec A-\cos A\right){\sec }^{2}A=\tan A$

Answer 1

L.H.S

$(\cos ecA-\sin A)(\sec A-\cos A){\sec }^{2}A$

We know that

$\sin \theta =\frac{1}{\cos ec\theta }$

$\cos \theta =\frac{1}{\sec \theta }$

$\tan \theta =\frac{1}{\cot \theta }$

Therefore,

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)\left(\frac{1}{{\cos }^{2}A}\right)$

$=\left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{1-{\cos }^{2}A}{\cos A}\right)\left(\frac{1}{{\cos }^{2}A}\right)$

We know that

${\cos }^{2}x=1-{\sin }^{2}x$

Therefore,

$=\left(\frac{{\cos }^{2}A}{\sin A}\right)\left(\frac{{\sin }^{2}A}{\cos A}\right)\left(\frac{1}{{\cos }^{2}A}\right)$

$=\left(\frac{\sin A}{\cos A}\right)$

$=\tan A$

Hence, proved.