Question

answer plzzzzzz ya fast

Answer 1

Given ,

3cot A = 4

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

=> $\frac{b}{p}$ = $\frac{4}{3}$ = k

therefore, b = 4k and p = 3k


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✌✌IF U ARE ASKED TO FIND SIN OR COS VALUE✌✌

APPLYING PYTHAGORAS THEOREM


${h}^{2}$ = ${p}^{2}$ + ${b}^{2}$

=> ${h}^{2}$ = ${(4k)}^{2}$ + ${(3k)}^{2}$

=> ${h}^{2}$ = ${16k}^{2}$ + ${9k}^{2}$

=> ${h}^{2}$ = ${25k}^{2}$

=> h = $\sqrt{ {25k}^{2}}$

=> h = 5k

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Now,

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

again,

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



$\huge \boxed { \frac{7}{25} }$