Question

If 3 cotA=4,check whether 1-tan^2A/1+tan^2A=cos^2A-sin^2A or not

Answer 1

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

Step-by-step explanation:

Given, 3 cot A= 4

$\Leftrightarrow cot A=\frac{4}{3}$

Then,  $tan A = \frac{3}{4}$

$cosecA=\sqrt{1+cot^2A}$

$\Leftrightarrow cosecA=\sqrt{1+(\frac{4}{3})^2 }$

$\Leftrightarrow cosecA=\sqrt{\frac{25}{9}$

$\Leftrightarrow cosecA={\frac{5}{3}$

$\Leftrightarrow sinA={\frac{3}{5}$

So,

$cos A=\sqrt{1- sin^2A}$

$\Leftrightarrow cos A=\frac{4}{5}$

L.H.S =$\frac{1-tan^2A}{1+tan^2A}$                          R.H.S=$cos^2A-sin^2A$

        $= \frac{1-(\frac{3}{4})^2 }{1+(\frac{3}{4})^2 }$                                   $=(\frac{4}{5})^2-(\frac{3}{5})^2$

       $=\frac{7}{25}$                                           $=\frac{7}{25}$

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

Answer 2

I hope it is helpful to you.❤