Question

If tan θ = a/b then prove that

a Sin θ - b Cos θ/a sin θ+ b cos θ=(a^2 - b^2)/ ( a^2+ b^2)​

Answer 1

Required Answer:-

Given:

• tan(x) = a/b

To prove:

$\rm \mapsto \dfrac{a \sin(x) - b \cos(x) }{a \sin(x) + b \cos(x) } = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

Proof:

Given that,

$\rm \implies \tan(x) = \dfrac{a}{b}$

$\rm \implies \dfrac{Height}{Base} = \dfrac{a}{b}$

Let Height = ka, Base = kb

Therefore Hypotenuse,

$\rm = \sqrt{ {(ka)}^{2} + (kb)^{2} }$

$\rm = \sqrt{ {k}^{2}( {a}^{2} +b^{2}) }$

$\rm = k\sqrt{ {a}^{2} +b^{2} }$

So,

$\rm \implies \sin(x) = \dfrac{ka}{k \sqrt{ {a}^{2} + {b}^{2} } } = \dfrac{a}{ \sqrt{ {a}^{2} + {b}^{2} } }$

And,

$\rm \implies \cos(x) = \dfrac{kb}{k \sqrt{ {a}^{2} + {b}^{2} } } = \dfrac{b}{ \sqrt{ {a}^{2} + {b}^{2} } }$

Taking LHS,

$\rm\dfrac{a \sin(x) - b \cos(x) }{a \sin(x) + b \cos(x) }$

$\rm = \dfrac{a \times \dfrac{a }{ \sqrt{ {a}^{2} + {b}^{2} } } - b \times \dfrac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } }{a \times \dfrac{a}{ \sqrt{ {a}^{2} + {b}^{2} } } + b \times \dfrac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } }$

$\rm = \dfrac{ \dfrac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } - \dfrac{ {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } }{ \dfrac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } + \dfrac{ {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } }$

$\rm = \dfrac{ \dfrac{ {a}^{2} - {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } }{ \dfrac{ {a}^{2} + {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } }$

$\rm = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

= RHS (Hence Proved)

Answer 2

Answer:

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