Question

If tan theta =4/3 find the value of 2sin theta-3cos theta/2sin theta+3cos theta

Answer 1

$\huge \red{Hello \: Friends}$

$\bf{ANS = = > }$

Given :

$\bf{tan \: theta = \frac{4}{3} }$

$\bf{To \: find : }$

$\huge\frac{2sin \: theta \: - \: 3cos \: theta}{2sin \: theta \: + \: 3cos \: theta}$

By Trignometry Identity,

$1 + tan {}^{2} theta\: = sec {}^{2} theta$

$1 + ( \frac{4}{3} ) {}^{2} = sec {}^{2} theta$

$1 + \frac{16}{9} = sec {}^{2} theta$

$\frac{9 + 16}{9} = sec {}^{2} theta$

$\frac{25}{9} = sec {}^{2} theta$

$sec \: theta = \sqrt{ \frac{25}{9} } = \frac{5}{3}$

$\boxed{sec \: theta = \frac{5}{3} }$

We know that,

$cos \: theta = \frac{1}{sec \: theta}$

$\boxed{cos \: theta = \frac{3}{5} }$

By Trignometry Identity,

$sin {}^{2} theta + cos {}^{2}theta = 1$

$sin {}^{2} theta + ( \frac{3}{5} ) {}^{2} = 1$

$sin {}^{2} theta \: = 1 - \frac{9}{25} = \frac{25 - 9}{25}$

$sin {}^{2} theta = \frac{16}{25}$

$sin \: theta \: = \sqrt{ \frac{16}{25} } = \frac{4}{5}$

$\boxed{sin \: theta = \frac{4}{5} }$

$\green{now : }$

$\frac{2sin \: theta \: - \: 3cos \: theta}{2sin \: theta \: + \: 3cos \: theta}$

$\huge \frac{2 \times \frac{4}{5 } - 3 \times \frac{3}{5} }{2 \times \frac{4}{5} + 3 \times \frac{3}{ 5} }$

$\huge \frac{ \frac{8}{5} - \frac{9}{5} }{ \frac{8}{5} + \frac{9}{5} }$

$\huge\frac{ \frac{8 - 9}{5} }{ \frac{8 + 9}{5} }$

$\huge \frac{ \frac{ - 1}{5} }{ \frac{17}{5} }$

$\huge \frac{ - 1}{17}$

$\boxed{ \boxed {ANS = = > \frac{ - 1}{17} }}$

$\huge \frak{Thanks}$

Answer 2

The value of $\bf \frac{2 \sin( \theta) -3 \cos( \theta)}{2\sin( \theta) + 3 \cos( \theta)}$ is -1/17.

Given:

• $\tan( \theta) = \frac{4}{3}$

To find:

• Value of $\frac{2 \sin( \theta) -3 \cos( \theta)}{2\sin( \theta) + 3 \cos( \theta)}$

Solution:

Formula to be used:

• $tan \theta = \frac{\text{side \: opposite \: to \: angle}}{\text{side \: adjacent\: to \: angle}}$
• $sin \theta = \frac{\text{side \: opposite \: to \: angle}}{hypotenuse}$
• $cos\theta = \frac{\text{side \: adjacent \: to \: angle}}{hypotenuse}$

Step 1:

Find hypotenuse of right triangle.

As,

Ratio of base to perpendicular is given so find hypotenuse.

i.e.

$\tan( \theta) =\frac{AB}{AC}= \frac{4k}{3k}$

Apply Pythagoras theorem

Hypotenuse² = Base² + Perpendicular² $CB^2= 9 {k}^{2} + 16 {k}^{2}$

Hypotenuse (CB)= 5k

Step 2:

Find sin theta form formula.

$\sin( \theta) = \frac{AB}{CB} =\frac{4k}{5k}$

or

$\bf \sin( \theta) = \frac{4}{5}$

Step 3:

Find cos theta from formula.

$\cos( \theta) = \frac{AC}{CB}= \frac{3k}{5k}$

or

$\bf \cos( \theta) = \frac{3}{5}$

Step 4:

Put the values in expression

$\frac{2 \sin( \theta) -3 \cos( \theta)}{2\sin( \theta) + 3 \cos( \theta)}=\frac{ 2 \times \frac{4}{5} - 3 \times \frac{3}{5}}{2 \times \frac{4}{5} + 3 \times \frac{3}{5}}$

or

$=\frac{ \frac{8}{5} - \frac{9}{5}}{\frac{8}{5} + \frac{9}{5} }$

or

$= \frac{-1}{17}$

Thus,

$\bf \frac{2 \sin( \theta) -3 \cos( \theta)}{2\sin( \theta) + 3 \cos( \theta)}= \frac{-1}{17}$

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