Question

4tanA=3 then find tha value of 4sinA-cosA/4sinA+cosA​

Answer 1

hey!!

tanA=3/4

Perpendicular=3

base=4

therefore ..

sinA=3/5

cosA=4/5

4sinA=4x3/5=12/5

cosA=4/5

12/5-4/5=8/5

answer is 8/5.

thanks!?

Answer 2

Concept: One of the most significant areas of mathematics is trigonometry, which has several applications in many different industries. The study of the right-angle triangle's sides, angles, and connection is the main focus of the field of mathematics known as "trigonometry." Therefore, it is helpful to use trigonometric formulas, functions, or trigonometric identities to find the unknown or missing angles or sides of a right triangle. The angles in trigonometry can either be expressed in degrees or radians. The most used trigonometric angles for computations include 0°, 30°, 45°, 60°, and 90° .

Given: 4tanA=3

To find: $\frac{4 sin A - cos A}{4 sin A + cos A}$

Solution:

As it is given

4 tan A = 3, therefore, $tan A = \frac{3}{4}$

We know that $tan = \frac{perpendicular}{base}$

So here perpendicular = 3k and base = 4k

First, we need to find out the hypotenuse using the Pythagoras theorem,

$H^2 = P^2 + B^2$  ;   where H is the hypotenuse,

                                      P is the perpendicular

                                      B is the base

Therefore,

$H^2 = (3k)^2 + (4k)^2\\ H^2 = 9k^2 + 16k^2\\H^2 = 25k^2\\ H = \sqrt{25k^2} \\H = 5k$

Now,$sin = \frac{perpendicular }{hypotenuse}$

Therefore, $sin A = \frac{3k}{5k}$

And $cos = \frac{base}{hypotenuse}$

Therefore, $cos A = \frac{4k}{5k}$

Now, according to question

4 sin A = $4 * \frac{3k}{5k} = \frac{12k}{5k}$

Putting the values in 4 sin A – cos A / 4 sin A + cos A, we get

$\frac{\frac{12k}{5k} - \frac{4k}{5k} }{\frac{12k}{5k} + \frac{4k}{5k} }\\\\\frac{\frac{8k}{5k} }{\frac{16k}{5k} } \\\\\frac{8k}{16k} \\\\\frac{1}{2}$

Hence, the value of 4 sin A – cos A / 4 sin A + cos A is $\frac{1}{2}$.

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