Question

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

Answer 1

this may help you my friend.

Answer 2

Answer:

yes,LHS=RHS

Step-by-step explanation:

If 3 cot A = 4,

$cot \: A = \frac{4}{3} \\ \\ tan \: A = \frac{3}{4}$

LHS

$\frac{1 - {tan}^{2} A}{1 + {tan}^{2} A} \\ \\ = \frac{1 - ({ \frac{3}{4}) }^{2} }{1 +( { \frac{3}{4} })^{2} } \\ \\ = \frac{1 - ({ \frac{9}{16}) } }{1 +( { \frac{9}{16} }) } \\ \\ = \frac{16 - 9}{16 + 9} \\ \\ = \frac{7}{25}$

If tan A= 3/4

then sin A=3/5 (from trigonometric functions)

cos A= 4/5

So,apply these values in RHS

${cos}^{2} A - {sin}^{2} A \\ \\ = ({ \frac{4}{5} })^{2} - ( { \frac{3}{5} })^{2} \\ \\ = \frac{16}{25} - \frac{9}{25} \\ \\ =\frac{16 - 9}{25} \\ \\ = \frac{7}{25}$

Hence RHS= LHS

yes,for the given value both LHS =RHS

Hope it helps you.