Question

if tan-1(4/3)=x,fine the value of cos x​

Answer 1

GIVEN :–

$\\\implies\bf\tan^{ - 1}(4/3) = x$

TO FIND :–

$\\\implies\bf Value \: \: of \: \: \cos(x) =?$

SOLUTION :–

$\\\implies\bf\tan^{ - 1}(4/3) = x$

$\\\implies\bf \tan \{\tan^{ - 1}(4/3) \}= \tan\{x\}$

$\\\implies\bf 4/3= \tan\{x\}$

$\\\implies\bf \tan(x) = \dfrac{4}{3}$

• We know that –

$\\\implies \large{ \boxed{\bf \tan(x) = \dfrac{ \sin(x) }{ \cos(x) }}}$

• So that –

$\\\implies\bf \dfrac{ \sin(x) }{ \cos(x) } = \dfrac{4}{3}$

• We also know that –

$\\\implies \large\bf{ \boxed{\sin^{2} ( \theta) + { \cos}^{2}( \theta) = 1}}$

• So that –

$\\\implies\bf \dfrac{ \sqrt{1 - \cos^{2} (x) } }{ \cos(x) } = \dfrac{4}{3}$

$\\\implies\bf 3\sqrt{1 - \cos^{2} (x)}=4\cos(x)$

• Square on both sides –

$\\\implies\bf 9(1 - \cos^{2} (x))=16\cos^{2} (x)$

$\\\implies\bf 9 -9\cos^{2} (x)=16\cos^{2} (x)$

$\\\implies\bf 9=16\cos^{2} (x) +9\cos^{2} (x)$

$\\\implies\bf 9=25\cos^{2} (x)$

$\\\implies\bf 25\cos^{2} (x) = 9$

$\\\implies\bf\cos^{2} (x) = \dfrac{9}{25}$

$\\\implies\bf\cos(x) = { \sqrt{\dfrac{9}{25}}}$

$\\\implies \large \red{ \boxed{\bf\cos(x) = \pm\dfrac{3}{5}}}$

Answer 2

GIVEN :–

$\\\implies\bf\tan^{ - 1}(4/3) = x$

TO FIND :–

$\\\implies\bf Value \: \: of \: \: \cos(x) =?$

SOLUTION :–

$\\\implies\bf\tan^{ - 1}(4/3) = x$

$\\\implies\bf \tan \{\tan^{ - 1}(4/3) \}= \tan\{x\}$

$\\\implies\bf 4/3= \tan\{x\}$

$\\\implies\bf \tan(x) = \dfrac{4}{3}$

• We know that –

$\\\implies \large{ \boxed{\bf \tan(x) = \dfrac{ \sin(x) }{ \cos(x) }}}$

• So that –

$\\\implies\bf \dfrac{ \sin(x) }{ \cos(x) } = \dfrac{4}{3}$

• We also know that –

$\\\implies \large\bf{ \boxed{\sin^{2} ( \theta) + { \cos}^{2}( \theta) = 1}}$

• So that –

$\\\implies\bf \dfrac{ \sqrt{1 - \cos^{2} (x) } }{ \cos(x) } = \dfrac{4}{3}$

$\\\implies\bf 3\sqrt{1 - \cos^{2} (x)}=4\cos(x)$

• Square on both sides –

$\\\implies\bf 9(1 - \cos^{2} (x))=16\cos^{2} (x)$

$\\\implies\bf 9 -9\cos^{2} (x)=16\cos^{2} (x)$

$\\\implies\bf 9=16\cos^{2} (x) +9\cos^{2} (x)$

$\\\implies\bf 9=25\cos^{2} (x)$

$\\\implies\bf 25\cos^{2} (x) = 9$

$\\\implies\bf\cos^{2} (x) = \dfrac{9}{25}$

$\\\implies\bf\cos(x) = { \sqrt{\dfrac{9}{25}}}$

$\\\implies \large \red{ \boxed{\bf\cos(x) = \pm\dfrac{3}{5}}}$