Question

if 4tanø=3,evaluate (4sinø-cosø=1)/(4 sinø+cosø-1)​

Answer 1

Given that : If tanθ = 3

$\sf : \; \implies tan \theta = \dfrac{3}{4}$

Need to find:

$\sf : \; \implies \dfrac{(4 sin \theta - cos \theta + 1 )}{(4 sin \theta + cos \theta - 1)}$

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❍  Let's say that, there's a right-angled triangle ΔABC which is right angled at B with one acute angle θ.

• Sides BC = 3k

                   AB = 4k

T H E R E F O R E,

• ~By applying Pythagoras theorem we get

⤳  AC² = AB² + BC²

⤳  AC² = [ 3k ]² + [ 4k ]²

⤳  AC² = 9k² + 16k²

⤳  AC² = 25k²

⤳  AC = √25k²

⤳  AC = 5k

⌑ We can write them as -

$\sf : \; \twoheadrightarrow tan \theta = \dfrac{BC}{AB} = \dfrac{3}{4}$

$\sf : \; \twoheadrightarrow sin \theta = \dfrac{BC}{AC} = \dfrac{3}{5}$

$\sf : \; \twoheadrightarrow \; cos \theta = \dfrac{AB}{AC} = \dfrac{4}{5}$

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Finding the value :-

$\sf : \; \implies \dfrac{(4 sin \theta - cos \theta + 1 )}{(4 sin \theta + cos \theta - 1)}$

$\sf : \; \implies \dfrac{ \bigg( 4 \times \dfrac{3}{5} - \dfrac{4}{5} + 1 \bigg)}{ \bigg( 4 \times \dfrac{3}{5} + \dfrac{4}{5} - 1 \bigg)}$

$\sf : \; \implies \dfrac{ \bigg( \dfrac{12}{5} - \dfrac{4}{5} + 1 \bigg)}{ \bigg( \dfrac{12}{5} + \dfrac{4}{5} - 1 \bigg)}$

$\sf : \; \implies \dfrac{ \bigg( \dfrac{12-4}{5} + 1 \bigg)}{ \bigg( \dfrac{12+4}{5} - 1 \bigg)}$

$\sf : \; \implies \dfrac{ \bigg( \dfrac{8}{5} + 1 \bigg)}{ \bigg( \dfrac{16}{5} - 1 \bigg)}$

$\sf : \; \implies \dfrac{ \bigg( \dfrac{8+5}{5} \bigg)}{ \bigg( \dfrac{16-5}{5} \bigg)}$

$\sf : \; \implies \dfrac{ \bigg( \dfrac{13}{5} \bigg)}{ \bigg( \dfrac{11}{5} \bigg)}$

$\sf : \; \implies \dfrac{13}{5} \div \dfrac{11}{5}$

$\sf : \; \implies \dfrac{13}{5} \times \dfrac{5}{11}$

$\boxed{\bf{ \bigstar \;\; \Rrightarrow \dfrac{13}{11}}}$

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•  Henceforth, the value is 3/11

Answer 2

Answer: 1/2

Explanation:

Here, 4 tan θ = 3

=> tan θ = 3/4

As tan A = sin A / cos A,

Therefore,

=> sin θ / cos θ = 3/4

=> 4 sin θ = 3 cos θ __(A)

Now we have to find:

(4 sin θ - cos θ) / (4 sin θ + cos θ)

= (3 cos θ - cos θ) / (3 cos θ + cos θ)

[From A]

= 2 cos θ / 4 cos θ

= 1/2.