Question

Question If 3 Cot A =4 then what is the value of​

Answer 1

Answer:

$\large\underline{\underline{\rm{\red{Given:-}}}}$

$\rm :\longmapsto 3cotA = 4$

$\large\underline{\underline{\rm{\red{To \ find :-}}}}$

$\rm :\longmapsto \dfrac{1-tan^2A}{1+tan^2A}$

$\large\underline{\underline{\rm{\red{Solution:-}}}}$

$\rm :\longmapsto 3cotA=4$

$\rm :\longmapsto cotA = \dfrac{4}{3}$

Hence, we can find tanA as:

$\rm :\longmapsto tanA = \dfrac{1}{cotA}$

$\rm :\longmapsto tanA = \dfrac{1}{\cfrac{4}{3}}$

$\rm :\longmapsto tanA = \dfrac{3}{4}$

Now putting this in the given expression,

$\rm :\longmapsto \dfrac{1-tan^A}{1+tan^2A} = \dfrac{1 - \left(\dfrac{3}{4}\right)^2}{1+\left(\dfrac{3}{4}\right)^2}$

$\rm :\longmapsto \dfrac{1-tan^A}{1+tan^2A} = \dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}$

$\rm :\longmapsto \dfrac{1-tan^A}{1+tan^2A} = \dfrac{\dfrac{7}{16}}{\dfrac{25}{16}}$

$\star\boxed{\boxed{\rm :\longmapsto \dfrac{1-tan^A}{1+tan^2A} = \dfrac{7}{25}}}$

$\large\underline{\underline{\rm{\red{More:-}}}}$

$\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}$

$\color{#FF7968}{\rule{36pt}{7pt}}\color{#4FB3F6}{\rule{36pt}{7pt}}\color{#FBBE2E}{\rule{36pt}{7pt}}\color{#60D399}{\rule{36pt}{7pt}}\color{#6D83F3}{\rule{36pt}{7pt}}$

Answer 2

$\sf \dag \: \red{Given :-}$

• 3Cot A = 4

$\sf \dag \: \red{To \: Find :-}$

$\boxed{ \sf \leadsto \orange{ \frac{1 - \tan ^{2}A }{1 + {tan}^{2}A } }}$

$\sf \dag \: \purple{Solution :-}$

$\sf \leadsto \: 3Cot A = 4 \\ \sf \leadsto \: cotA = \frac{4}{3} \: \: \:$

We know that,

$\sf \leadsto \: tan \: A = \frac{1}{cot \:A }$

$\sf \therefore \: tan \: A = \frac{1}{ \frac{4}{3} } = \frac{3}{4}$

Now we will put the derived value of tan A in the given expression, we will get,

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

$\boxed{ \orange \bigstar \: \pink{ \sf \frac{1 - {tan}^{2} A}{1 + {tan}^{2}A } = \frac{7}{25} }}$