Question

if secA=25/7 then find 3tanA-tan3A/1-3tan2A​

Answer 1

Answer:

ans is 32 so right these ans ok frnd

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

sec A = 25/7

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure of right angled triangle according to the question;

As we know that;

$\boxed{\bf{sec\:\theta = \frac{Hypotenuse}{Base} }}$

$\mapsto\tt{sec\:A = \dfrac{BC}{AB} =\dfrac{25}{7} }$

$\underline{\mathcal{USING\:BY\:PYTHAGORAS\:THEOREM\::}}$

$\longrightarrow\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }$

$\longrightarrow\sf{(BC)^{2} = (AB)^{2} + (AC)^{2} }$

$\longrightarrow\sf{(25)^{2} = (7)^{2} + (AC)^{2} }$

$\longrightarrow\sf{625 = 49 + (AC)^{2} }$

$\longrightarrow\sf{(AC)^{2} = 625 - 49 }$

$\longrightarrow\sf{(AC)^{2} = 576 }$

$\longrightarrow\sf{AC =\sqrt{576} }$

$\longrightarrow\bf{AC = 24\:cm }$

Now,

$\mapsto\tt{\dfrac{3tan\:A - tan^3A}{1-3tan^2A}}$

∴ tan Ф = perpendicular/base

$\mapsto\tt{\dfrac{3 \times \bigg(\dfrac{AC}{AB}\bigg) - \bigg(\dfrac{AC}{AB}\bigg)^{3 } }{1-3\times \bigg(\dfrac{AC}{AB} \bigg)^{2}}}$

$\mapsto\tt{\dfrac{3 \times \bigg(\dfrac{24}{7}\bigg) - \bigg(\dfrac{24}{7}\bigg)^{3 } }{1-3\times \bigg(\dfrac{24}{7} \bigg)^{2}}}$

$\mapsto\tt{\dfrac{ \dfrac{72}{7}- \dfrac{13824}{343} }{1-3\times \dfrac{576}{49}}}$

$\mapsto\tt{\dfrac{ \dfrac{3528-13824}{343} }{1-\dfrac{1728}{49}}}$

$\mapsto\tt{\dfrac{ \dfrac{3528-13824}{343} }{\dfrac{49 - 1728}{49}}}$

$\mapsto\tt{\dfrac{ - \dfrac{10296}{343} }{-\dfrac{1679}{49}}}$

$\mapsto\tt{-\dfrac{10296}{\cancel{343}} \times -\dfrac{\cancel{49}}{1679} }$

$\mapsto\tt{\dfrac{10296}{7\times 1679} }$

$\mapsto\tt{\dfrac{10296}{11753} }$