CosA 0.6,show that(5sinA-3tanA)=0

Given, cos{tex}\theta{/tex} = 0.6 {tex}= \frac { 6 } { 10 } = \frac { 3 } { 5 }{/tex}
Let us draw a triangle ABC in which {tex}\angle{/tex}B = 90°.

Let {tex}\angle{/tex}A = {tex}\theta{/tex}°.
Then, {tex}\cos \theta = \frac { A B } { A C } = \frac { 3 } { 5 }{/tex}
Let AB = 3k and AC = 5k, where k is positive.
By Pythagoras' theorem, we have
AC² = AB² + BC²
{tex}\Rightarrow{/tex}BC² = AC² - AB² 
= (5k)² - (3k)² = 25k² - 9k² = 16k²
{tex}\Rightarrow \quad B C = \sqrt { 16 k ^ { 2 } } = 4 k{/tex}
{tex}\sin \theta = \frac { A B } { A C } = \frac { 4 k } { 5 k } = \frac { 4 } { 5 }{/tex}
{tex}\cos \theta = \frac { 3 } { 5 }{/tex}
{tex}\tan \theta = \frac { \sin \theta } { \cos \theta } = \left( \frac { 4 } { 5 } \times \frac { 5 } { 3 } \right) = \frac { 4 } { 3 }{/tex}
{tex}\Rightarrow ( 5 \sin \theta - 3 \tan \theta ) = \left( 5 \times \frac { 4 } { 5 } - 3 \times \frac { 4 } { 3 } \right) = 0{/tex}
Hence, (5sin{tex}\theta{/tex} - 3 tan{tex}\theta{/tex}) = 0.