`5 sin theta+ 2 cos theta = 5` (given `tan 21^alpha 48'=0.4`)

Solve: 5sintheta + 2cos theta=5 , given tan21^(@) 48'=2/5

If 4 sin theta + 3 cos theta =5 then tan theta =

If theta lies in the first quadrant and 5 tan theta = 4 , " then " (5 sin theta - 3 cos theta )/( sin theta + 2 cos theta )= .............. A) 0 B) 1 C) 1/6 D) 5/14

If 0 lt theta lt (pi)/2 and 5 tan theta = 4 then (5 sin theta - 3 cos theta) / (sintheta +2 cos theta ) = 5/14

If 5 tan theta =4 , then frac(5sin theta - 3cos theta )(5 sin theta + 2cos theta ) is equal to

If theta lies in the first quadrant and 5 tan theta = 4 , " then " (5 sin theta - 3 cos theta )/( sin theta + 2 cos theta )=

If theta lies in the first quadrant and 5 tan theta = 4 , " then " (5 sin theta - 3 cos theta )/( sin theta + 2 cos theta )=

If 5 tan theta = 4 , show that ( 5 sin theta - 3 cos theta)/(5 sin theta + 3 cos theta) = (1)/(7)

If 5 tan theta = 4 , show that ( 5 sin theta - 3 cos theta)/(5 sin theta + 3 cos theta) = (1)/(7)