Question

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

Answer 1

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...

Answer 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 = $\dfrac{a}{b}$

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

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

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

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

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

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

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

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

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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.