Question

please answer this question ​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Given :

$3\:Cot \theta = \dfrac{3(B)}{4(P)}$

Here base is 3 and perpendicular is 4

By using Pythagoras theorem :-

$\bold{\boxed{{H}^{2} = {P}^{2} + {B}^{2}}}$

${H}^{2} = {4}^{2} + {3}^{2}$

${H}^{2} = 16 + 9 = 25$

$H= \sqrt{25} = 5$

Taking L.H.S

=>we know that tan is perpendicular /base

$= > \dfrac{(1 - {tan}^{2} \theta)}{(1 + {tan}^{2} \theta)}$

$= > \dfrac{1 - {(\dfrac{3}{4} )}^{2} }{1 + { (\dfrac{3}{4} )}^{2} }$

$= > \dfrac{1 - \dfrac{9}{16} }{1 + \dfrac{9}{16} }$

$= > \dfrac{ \dfrac{16 - 9}{16} }{ \dfrac{16 + 9}{16} }$

$= > {\red{\dfrac{7}{25}}}$

Now,taking R.H.S =>

$\bold{= > {cos}^{2} \theta - {sin}^{2} \theta}$

=>Cos is Base /hypotenuse and

=> sin is perpendicular/base

$= > {( \dfrac{4}{5}) }^{2} - { (\dfrac{3}{5} )}^{2}$

$= > \dfrac{16}{25} - \dfrac{9}{25} = {\red{\dfrac{7}{25} }}$

Hence,proved L.H.S=R.H.S

Answer 2

ok, thanks for information