Question

If tan theta =a/b, prove that a sin theta- b cos theta / a sin theta - b cos theta=a^2-b^2/a^2+b^2
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Answer 1

Note : Theta is written as A.

Given value of tanA is a / b, and from the properties of trigonometry we know that the value of tangent in terms of cosine and sine is cosine / sine.

TanA = sinA / cosA

According to the question : -

= > tanA = $\dfrac{a}{b}$

= > tanA =$\sf \dfrac{sinA}{cosA}=\dfrac{a}{b}$ { from trigonometric properties }

= > =$\sf \dfrac{sinA}{cosA}=\dfrac{a}{b}$

Multiplying both sides by a / b.

$= > \sf \dfrac{ a }{ b} \times \dfrac{ sinA }{ cosA }= \dfrac{a }{b } \times \dfrac{a }{ b} \\ \\ </p><p>= > \sf \dfrac{ asinA }{bcosA} = \dfrac{ a^2 }{ b^2 }$

Applying Componendo & Dividendo : -

Componendo & Dividendo : A term from ratio and proportional, according to this, if a fraction is equal to an another fraction{ let a / b is equal to x / y }, so ( a + b ) / ( a - b ) = ( x + y ) / ( x - y ).

Thus,

$\implies \sf \dfrac{ ( a sinA + b cosA ) }{( a sinA - b cosA )} = \dfrac{ ( a^2 + b^2 )}{ ( a^2 - b^2 )} \\ \\ \\ \implies \sf \dfrac{ ( a^2 - b^2 )}{ ( a^2 + b^2 )} = \dfrac{ ( a sinA - b cosA ) }{( a sinA + b cosA )} =$

Hence proved.

Answer 2

Solution :-

tan θ = a/b

sinθ/cosθ = a/b------(1)

LHS = (a sinθ- b cos θ)/(a sinθ + b cosθ)

= (asinθ/cosθ - bcosθ/cosθ)/(asinθ/cosθ+ bcosθ/cosθ)

dividing each term with cosθ

=(asinθ/cosθ-b/1)/(asinθ/cosθ+b/1)

=(a*a/b-b/1)/(a*a/b +b/1)

[from (1)]

= (a^2-b^2)/(a^2+b^2)

= RHS