Question

4sinα+3cosα=5 then tanα=?

Answer 1

4sinα+3cosα= 5

dividing the both sides by cosα

$\frac{4sin \alpha }{cos \alpha }+\frac{3cos \alpha }{cos \alpha }= \frac{5}{cos \alpha }$

${4tan \alpha }+{3}= \frac{5}{cos \alpha }$

squaring

16tan²α + 9 + 24tanα = 25tan²α + 25

9tan²α - 24tanα + 16 = 0

(3tanα)² - 2 . 3tanα .4 + 4² = 0

(3tanα - 4)² = 0

tanα = 4/3 ANSWER 

Answer 2

4sina +3cosa = 5
or sina=(5-3cosa)/4
sina*sina=((5-3cosa)/4)((5-3cosa)/4)=(25+9(cosa*cosa)-30cosa)/16

But sina*sina + cosa*cosa =1
So (25 +9(cosa*cosa)-30cosa)/16 +cosa*cosa =1
Simplifying we get 25(cosa*cosa) -30cosa + 9 =0
or 25(cosa*cosa) -15cosa -15cosa +9 =0
or  5cosa(5cosa - 3) -3(5cosa - 3) = 0
or (5cosa - 3)(5cosa -3) = 0
or 5cosa -3 = 0
or cosa = 3/5
So sina = (5 - 3(3/5))/4 = 4/5
So tana=sina/cosa = (4/5)/(3/5) = 4/3