Question

If 3cotA=4,then check whether 1-tan²A/1+tan²A=cos²A-sin²A

Answer 1

hope this helps you............

Answer 2

LHS = RHS

Step-by-step explanation:

Given

$3 \cot(a) = 4 \\ \\ and \: prove \\ \\ \frac{1 - { \tan }^{2}a }{1 + { \tan }^{2}a} = { \cos }^{2} a - { \sin }^{2} a$

now,

Draw a right angle triangle ΔABC

• right angle at A
• as shown in the adjoining figure in above pic

as we know given

$\cot(a) = \frac{4}{3}$

let AB =4k

AC = 3k

where K is some positive real number..

now by usinh Pythagoras theorem

BC² =AB² +AC²

put the value we get

BC² = (4K)² +(3K)²

= 16 K² +9 K²

= 25K²

taking squre root out we get..

BC = 5k

Now

$\tan(a) = \frac{ac}{ab} \\ \\ \: \ \: \: \: \: \: \: \: \: \: \: \: \: = \frac{3k}{4k} = \frac{3}{4} \\ \\ \sin(a) = \frac{ac}{bc} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{3k}{5k} = \frac{3}{5} \\ \\ \cos(a) = \frac{ab}{bc} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = \frac{4k}{5k} = \frac{4}{5}$

now as we have

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

take L.H.S AND R.H.S

LHS =

$\frac{1 - { \tan }^{2}a }{1 + { \tan }^{2}a} \\ \\ put \: all \: the \: values \: we \: get \\ \\ \frac{1 - ( \ { \frac{3}{4} }^{2} ) }{1 + ( \frac{ {3}^{2} }{4} )} \\ \\ \frac{1 - \frac{9}{16} }{1 + \frac{9}{16} } \\ \\ \frac{ \frac{16 - 9}{16} }{ \frac{16 + 9}{16} } \\ \\ \\ \frac{7}{16} \times \frac{16}{25} \: \: \\ reciprocl \: the \: fraction$

L.HS =

$\frac{7}{25}$

now take RHS

${ \ \cos(a) }^{2} - { \sin(a) }^{2} \\ \\ put \: the \: value \: we \: get \\ \\ {1}^{2} - ( { \frac{3}{5} }^{2} ) \\ \\ 1 - \frac{9}{25} \\ \\ \frac{25 - 9}{25} \\ \\ \frac{7}{25}$

there for

we get LHS = RHS

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