Question

​if 5tan tita = 4, find the value of 5sin tita- 3 cos tita+5sin tita -2cos tita​

Answer 1

$\bf \LARGE \it Hey \: User!!!$

given :-

5tan∅ = 4

therefore tan∅ = 4/5

consider any right angle ∆abc,

where AB is the base, BC is the perpendicular and AC is the hypotenuse.

we know that tan∅ = perpendicular/base

$\rm \small \therefore \: perpendicular = 4 \: and \: base = 5$
by Pythagoras theorem we get :-

>> AC² = AB² + BC²
>> AC² = 4² + 5²
>> AC² = 16 + 25
>> AC² = 41
>. AC = √41

▶sin∅ = 4/√41
▶cos∅ = 5/√41

$\therefore \rm \small 5\sin \theta - 3 \cos \theta + 5 \sin \theta - 2 \cos \theta \\ \rm \tiny = 5 \times \frac{4}{ \sqrt{41} } - 3 \times \frac{5}{ \sqrt{41} } + 5 \times \frac{4}{ \sqrt{41} } - 2 \times \frac{5}{41} \\ \tiny = \frac{20}{ \sqrt{41} } - \frac{15}{ \sqrt{41} } + \frac{20}{ \sqrt{41} } - \frac{10}{ \sqrt{41} } \\ \tiny = \boxed{ \frac{15}{ \sqrt{41} } } \rm \small \: final \: answer$

$\bf \LARGE \it Cheers!!!$