Question

If tan A =
4/3
then find the values of the following:
i) sin 2A
ii) cos 2A
iii) tan 2A
iv) cot 2A​

Answer 1

Step-by-step explanation:

tan a = 4/3 , we know

tan 2a = 2tan a / (1 - tan^2 a)

= 2 *4/3/{(1 - 16/9)}

= (8/3)/(-7/9)

= - 24/7

Answer 2

S O L U T I O N :

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

tan A = 4/3

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

Firstly, attachment a figure of right angle triangle according to the question.

As we know that;

$\boxed{\bf{tan\:\theta = \frac{Perpendicular}{Base} }}$

$\mapsto\tt{tan\:A = \dfrac{AC}{AB} =\dfrac{4}{3} }$

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

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

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

$\longrightarrow\sf{(BC)^{2} = (3)^{2} + (4)^{2}}$

$\longrightarrow\sf{(BC)^{2} = 9 + 16}$

$\longrightarrow\sf{BC =\sqrt{25} }$

$\longrightarrow\bf{BC =5\:unit }$

Now,

sin² A

$\mapsto\tt{sin^{2}A = \dfrac{Perpendicular}{Hypotenuse} }$

$\mapsto\tt{sin^{2}A = \bigg(\dfrac{AC}{BC}\bigg)^{2} }$

$\mapsto\tt{sin^{2}A = \bigg(\dfrac{4}{5}\bigg)^{2} }$

$\mapsto\tt{sin^{2}A = \dfrac{16}{25}}$

cos² A

$\mapsto\tt{cos^{2}A = \dfrac{Base}{Hypotenuse} }$

$\mapsto\tt{cos^{2}A = \bigg(\dfrac{AB}{BC}\bigg)^{2} }$

$\mapsto\tt{sin^{2}A = \bigg(\dfrac{3}{5}\bigg)^{2} }$

$\mapsto\tt{sin^{2}A = \dfrac{9}{25} }$

tan² A

$\mapsto\tt{tan^{2}A = \dfrac{Perpendicular}{Base} }$

$\mapsto\tt{tan^{2}A = \bigg(\dfrac{AC}{AB}\bigg)^{2} }$

$\mapsto\tt{tan^{2}A = \bigg(\dfrac{4}{3}\bigg)^{2} }$

$\mapsto\tt{tan^{2}A = \dfrac{16}{9} }$

cot² A

$\mapsto\tt{cot^{2}A = \dfrac{Base}{Perpendicular} }$

$\mapsto\tt{cot^{2}A = \bigg(\dfrac{AB}{AC}\bigg)^{2} }$

$\mapsto\tt{cot^{2}A = \bigg(\dfrac{3}{4}\bigg)^{2} }$

$\mapsto\tt{cot^{2}A = \dfrac{9}{16} }$