Math, asked by BrainlyHelper, 1 year ago

If tan\Theta =\frac{a}{b}, prove that \frac{asin\Theta+bcos\Theta}{asin\Theta+bcos\Theta} =\frac{a^{2}-b^{2}  }{a^{2}+b^{2}  }

Answers

Answered by nikitasingh79
0

SOLUTION IS IN THE ATTACHMENT.

** Trigonometry is the study of the relationship between the sides and angles of a triangle.

The ratio of the sides of a right angled triangle with respect to its acute angles are called trigonometric ratios.

** For any acute angle in a right angle triangle the side opposite to the acute angle is called a perpendicular(P),  the side adjacent to this acute angle is called the base(B) and side opposite to the right angle is called the hypotenuse(H).

** Find the third  side of the right ∆ ABC by using Pythagoras theorem (AC² = AB² + BC²).

HOPE THIS ANSWER WILL HELP YOU...

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Answered by abhi569
2

Hello there !

Note : Theta is written as A.


Your question needs a correction.

Correct question : If\: tan\Theta =\dfrac{a}{b},\\ prove \:that\: \frac{asin\Theta-bcos\Theta}{asin\Theta+bcos\Theta} =\frac{a^{2}-b^{2}}{a^{2}+b^{2}}



Given, tan A = \dfrac{a}{b}

-----------------------------------------------------------

We know that tanA is the ratio of height of the triangle for angle A to the base of the triangle for angle A.Mathematically we can say that tanA = \mathsf{\dfrac{height}{base}}


Now,

tanA = \mathsf{\dfrac{height}{hypotenuse}}

tanA = \mathsf{\dfrac{\dfrac{height}{hypotenuse}}{\dfrac{base}{hypotenuse}}}

-------------------------------------

\dfrac{height}{hypotenuse}= sinA\:\:\:\:\:\:\:\bold{from\:trigonometric\:identities}


\dfrac{base}{hypotenuse}= cosA\:\:\:\:\:\:\:\bold{from\:trigonometric\:identities}

---------------------------------------


So, tanA = \dfrac{sinA}{cosA}


-----------------------------------------------------------



Solving right hand side,  

\dfrac{asin\Theta-bcos\Theta}{asin\Theta+bcos\Theta}


Divide both sides by cosA.


\dfrac{a \bigg(\dfrac{sinA}{cosA} \bigg)-b\bigg(\dfrac{cosA}{cosA} \bigg)}{a \bigg(\dfrac{sinA}{cosA}\bigg) +b\bigg( \dfrac{cosA}{cosA} \bigg)}


\dfrac{a ( tanA )-b(1) }{a (tanA) +b( 1 )}


tanA = \dfrac{a}{b}              [ Given ]


\dfrac{a\bigg( \dfrac{a}{b}\bigg) - b }{ a\bigg( \dfrac{a}{b} \bigg)+ b }


\dfrac{\dfrac{a^2 - b^2 }{b}}{\dfrac{a^2 + b^2}{b}}


\dfrac{a^2 - b^2}{a^2 + b^2}



Hence, proved that If  tan\Theta =\dfrac{a}{b}\\ then\: \frac{asin\Theta-bcos\Theta}{asin\Theta+bcos\Theta} =\frac{a^{2}-b^{2}  }{a^{2}+b^{2}}



Method 2. this method is very easy if you know componendo and dividendo { ratio & proporation }


Given, tanA = \dfrac{a}{b}


We know that tanA is the ratio of height of the triangle for angle A to the base of the triangle for angle A.Mathematically we can say that tanA = \mathsf{\dfrac{height}{hypotenuse}}


Now,

tanA = \mathsf{\dfrac{height}{hypotenuse}}

tanA = \mathsf{\dfrac{\dfrac{height}{hypotenuse}}{\dfrac{base}{hypotenuse}}}

-------------------------------------

\dfrac{height}{hypotenuse}= sinA\:\:\:\:\:\:\:\bold{from\:trigonometric\:identities}


\dfrac{base}{hypotenuse}= cosA\:\:\:\:\:\:\:\bold{from\:trigonometric\:identities}

---------------------------------------


So, tanA = \dfrac{sinA}{cosA}

---------------------------------------------------


Then,

\dfrac{sinA}{cosA}=\dfrac{a}{b}


Multiply by a on both sides ( numerator ) and by b on both sides ( denominator )


\dfrac{a sinA}{b cosA}= \dfrac{a^2}{b^2}


By Componendo and dividendo


\dfrac{a sinA +b cosA}{a sinA- bcosA }= \dfrac{a^2 + b^2}{a^2 - b^2}


\dfrac{a^2 - b^2}{a^2+b^2} =\dfrac{a sinA -bcosA}{a sinA + bcosA}



Hence, proved.

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