Question

Prove the following :

$\frac{\tan ({90}^{\circ }-A)\cot A}{{\csc }^{2}A}-{\cos }^{2}A=0.$

Answer 1

L.H.S$=\frac{\tan ({90}^{\circ }-A)\cot A}{{\csc }^{2}A}-{\cos }^{2}A$

We know that,

$\cos (90-\theta )=\sin \theta$

$\sin (90-\theta )=\cos \theta$

$\tan (90-\theta )=\cot \theta$

$\cot \theta =\frac{\cos \theta }{\sin \theta }$

$\csc \theta =\frac{1}{\sin \theta }$

$=\frac{\cot \left(A\right)\cot A}{{\csc }^{2}A}-{\cos }^{2}A$

$=\frac{{\cot }^{2}A}{{\csc }^{2}A}-{\cos }^{2}A$

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

$={\cos }^{2}A-{\cos }^{2}A$

$=0$=R.H.S

$\because \frac{\tan ({90}^{\circ }-A)\cot A}{{\csc }^{2}A}-{\cos }^{2}A=0$

hence proved