Question

If $tan\Theta=\frac{4}{5}$, find the value of $\frac{cos\Theta-sin\Theta}{cos\Theta+sin\Theta}$

Answer 1

SOLUTION :  

Given : tan θ = ⅘  

We have to find the value of :  (cos θ - sin θ)/(cos θ + sin θ)

= (cos θ/cos θ - sin θ/cos θ)/(cos θ/cos θ + sin θ/cos θ)

[on dividing both the numerator and denominator by cosθ]

= (1 - sinθ/cosθ) / (1 + sinθ/cosθ)

= ( 1 - tan θ) / (1 + tan θ)

[By using identity , sinθ/cosθ = tan θ]

= (1 - ⅘) / (1 + ⅘)

= [(5 - 4)/5 ] /[ (5 + 4)/5]

= (⅕) /( 9/5)

= ⅕ × 5/9

= 1/9  

Hence , the value of (cos θ - sin θ)/(cos θ + sin θ) is 1/9 .

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Answer 2

Answer:

$\frac{cos\Theta-sin\Theta}{cos\Theta+sin\Theta}=\frac{1}{9}$

Solution:

If $tan\Theta=\frac{4}{5}$

that means in a right triangle the ratio of perpendicular to base is 4/5=k

perpendicular = 4k

Base = 5k

Hypotenuse=

$\sqrt{16 {k}^{2} + 25 {k}^{2} } \\ \\ = \sqrt{41 {k}^{2} } \\ \\ = k \sqrt{41}$
So,

$sin\Theta=\frac{4}{\sqrt{41}}$

$cos\Theta=\frac{5}{\sqrt{41}}$

So,

$\frac{cos\Theta-sin\Theta}{cos\Theta+sin\Theta} = \frac{ \frac{5}{ \sqrt{41} } - \frac{4}{ \sqrt{41} } }{ \frac{5}{ \sqrt{41} } + \frac{4}{ \sqrt{41} } } \\ \\ = \frac{5 - 4}{ \sqrt{41} } \times \frac{ \sqrt{41} }{4 + 5} \\ \\ = \frac{1}{9}$