Question

if tanΘ=m/n then prove that mSinΘ-nCosΘ/mSinΘ+nCosΘ=m²-n²/m²+n²

Answer 1

Step-by-step explanation:

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Answer 2

Let : I have taken value of angle as "x" in place of theta.

Given :

• $\tan(x) = \dfrac{m}{n}$

To prove :

$\dfrac{m \sin(x) - n \cos(x) }{m \sin(x) + n \cos(x) } = \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} }$

Identity used :

• tan ∅ = sin ∅ ÷ cos ∅

Solution :

$\bullet \: \tan(x) = \dfrac{m}{n} \\ \\ \bullet \: \dfrac{ \sin(x) }{ \cos(x) } \: = \: \dfrac{m}{n} \\ \\ \bullet \: \sin(x) = \: ( \dfrac{m}{n} ) \cos(x)$

Now ,

$LHS \implies \: \dfrac{m \sin(x) - n \cos(x) }{m \sin(x) + n \cos(x) } \\ \\ putting \: value \: of \: \sin(x) \: in \: above \: we \: get \\ \\ \implies \dfrac{m \times (\dfrac{m}{n} \cos(x) ) - n \cos(x) }{m \times (\dfrac{m}{n} \cos(x)) + n \cos(x)} \\ \\ \\ \implies \: \dfrac{ \dfrac{ {m}^{2} \cos(x) }{n} - n \cos(x) }{ \dfrac{ {m}^{2} \cos(x) }{n} + n \cos(x) } \\ \\taking \: lcm \\ \implies \: \dfrac{ \dfrac{ {m}^{2} \cos(x) \: - n \times n \cos(x) }{n} }{ \dfrac{ {m}^{2} \cos(x) + n \times n \cos(x) }{n} }$

Taking , ( cos (x) ÷ n ) common from numerator & denominator

we get;

$\implies \: \dfrac{ \dfrac{ \cos(x) }{n} ( {m}^{2} - {n}^{2} )}{ \dfrac{ \cos(x) }{n}( {m}^{2} + {n}^{2}) } \\ \\ \dfrac{ \cos(x) }{n} \: get \: cancelled \\ \\ \implies \: \dfrac{( {m}^{2} - {n}^{2} ) }{( {m}^{2} + {n}^{2}) } \: = \: RHS$

Hence PROVED