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If 5tan theta =4,find the value of 5sin theta -3cos theta by 5sin theta + 2cos theta

If 5tan theta =4,find the value of 5sin theta -3cos theta by 5sin theta + 2cos theta

Grade:11

9 Answers

Meet
137 Points
3 years ago
Let theta be @ for simpler in writing , then divide whole by cos@ then the new formed is as (5tan@-3)/(5tan@+2)= 1/6
Arun
25763 Points
3 years ago
if 5 tan \theta = 4
then
tan\theta = 4/5
 
now
(5 sin\theta – 3cos\theta) / (5 sin\theta + 2cos\theta)
divide by cos \theta in both numerator and denominator
(5 tan\theta – 3) / (5 tan\theta + 2)
 put the value of tan \theta
(4 – 3) / (4 + 2)
 = 1/6
 
Regards
Arun (askIITians forum expert)
Samaresh
19 Points
3 years ago
If I done today is equal to 4 then find the value of sin theta minus 3 cos theta by sin theta + 2 cos theta
Priyanshu priyadarshi
300 Points
3 years ago
5 tan theta = 4Tan theta = 4/5We know that tan theta = p/b so, here 4 is perpendicular and 5 is baseH=√square of 5 + square of 4H= √25+16H=√41a.t.q = 5 sin theta -3 cos theta / 5 sin theta + 2 cos theta =5 × 4/√41 - 3 × 5/41 / 5× 4/√41 + 2× 5/√4120/√41 - 15/√41 / 20/√41 + 10/√415/√41 / 30/√415 /√41 × √41 /30So , √41 will be cutted nowTherefore 5/30 remains Now 5/30 becomes 1/6 by cutting both.If my answer is helpful fir you so pls like it
Soumendu Majumdar
159 Points
3 years ago
Dear Student,
Given: 5tan theta = 4 implies tan theta = 4/5
Now ( 5sin theta – 3cos theta )/( 5sin theta + 2cos theta )
= ( 5tan theta – 3 )/( 5tan theta + 2 )   by dividing numerator and denominator by cos theta
= ( 4 – 3 )/( 4 + 2)   by putting the value of tan theta = 4/5
= 1/6
 
Hope it helps!
Ludicrous Lunatic
53 Points
3 years ago
If 5 tan theta is 4 then sin theta is 4/5. Sin theta will (41^1/2)/
Samuel Garry
61 Points
3 years ago
 
if 5 tan \theta = 4
then
tan\theta = 4/5
 
now
(5 sin\theta – 3cos\theta) / (5 sin\theta + 2cos\theta)
divide by cos \theta in both numerator and denominator
(5 tan\theta – 3) / (5 tan\theta + 2)
 put the value of tan \theta
(4 – 3) / (4 + 2)
 = 1/6
 
Neha
17 Points
3 years ago
According to the question
5tan theta =4
Then,tan theta =4/5
Draw a triangle ABC ∆,90^°at angle B&angleC is theta, such that side AB is the perpendicular,BC the base &AC be the hypotenuse
Then we know ,tan theta=perpendicular/base=AB/BC
Now let AB=4k&BC=5k
In ∆ABC
AC²= AB² + BC²
AC²=4k²+5k²
AC²=16k²+25k²
AC=√41k
Sin theta=perpendicular/hypotenuse=4k/√41k=4/√41
Cos theta = base/perpendicular=5k/√41k=5/√41
5sin theta-3cos theta/5sin theta-2cos theta
5×4/√41-3×5/√41/5×4/√41+2×5√41
=5/√41×√41/30
=1/6
 
 
 
Durwa Thakur
23 Points
3 years ago
 
Hey there!
Here's the solution to your question.
According to the question
5tan theta =4
Then,tan theta =4/5
Draw a triangle ABC ∆,90^°at angle B&angleC is theta, such that side AB is the perpendicular,BC the base &AC be the hypotenuse
Then we know ,tan theta=perpendicular/base=AB/BC
Now let AB=4k&BC=5k
In ∆ABC
AC²= AB² + BC²
AC²=4k²+5k²
AC²=16k²+25k²
AC=√41k
Sin theta=perpendicular/hypotenuse=4k/√41k=4/√41
Cos theta = base/perpendicular=5k/√41k=5/√41
5sin theta-3cos theta/5sin theta-2cos theta
5×4/√41-3×5/√41/5×4/√41+2×5√41
=5/√41×√41/30
=1/6.
Cheers.
 

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