Question

If 4 tan theta = 3 , then find (4 sin theta - cos theta / 4 sin theta + cos theta)

Answer 1

Answer:-

Given:-

$4 \: tan\theta=3$

To find:-

$\frac{4 \: sin\theta-cos\theta}{4 \: sin\theta+cos\theta} =??$

Solution:-

⇒ $4 \: tan\theta=3$

⇒ $tan\theta=\frac{3}{4}............................eq \: 1$

⇒ $\frac{4 \: sin\theta-cos\theta}{4 \: sin\theta+cos\theta} = \frac{4\frac{sin\theta}{cos\theta}-1}{4\frac{sin\theta}{cos\theta}+1 }$      [divide by cosθ in both numerator and dinominator]

⇒ $\frac{4 \: tan\theta-1}{4 \: tan\theta+1}$

                           $[tan\theta=\frac{sin\theta}{cos\theta} ]$

⇒ $\frac{4(\frac{3}{4})-1 }{4(\frac{3}{4})+1 }$

⇒ $\frac{3-1}{3+1}$

⇒ $\frac{2}{4} =\frac{1}{2}$

                            [put the value from eq 1]

Hence, the value of $\frac{4 \: sin\theta-cos\theta}{4 \: sin\theta+cos\theta}$ is $\frac{1}{2}$

Answer 2

Answer:

Step-by-step explanation:

4tanA = 3 ---( 1 )

tanA = 3/4

=> tan² A = 9/16

=> sec²A - 1 = 9/16

=> sec²A = 9/16 + 1

=> sec²A = ( 9 + 16)/16

=> Sec²A = 25/16

=> SecA = 5/4 ---( 2 )

Now ,

(4sinA-cosA+1)/(4sinA+cosA-1)

Divide numerator and denominator

With cosA, we get

(4tanA-1+secA)/(4tanA+1-secA)

= (3-1+5/4)/(3+1-5/4) [from(1)&(2)]

= (2+5/4)/(4-5/4)

Divide numerator and denominator

With 4 , we get

= ( 8 + 5 )/( 16 - 5 )

= 13/11