Question

tell the answer please?​

Answer 1

It is given that tanθ = a/b

LHS = a sinθ - b cosθ / a sinθ + b cosθ

Dividing the numerator and denominator by cosθ, we get : a tanθ - b / a tanθ + b {since, tanθ = sinθ / cosθ}

Now substituting the value of tanθ in the above expression, we get :

a (a/b) - b / a (a/b) + b

= a²/b - b / a²/b + b

= a²-b² / a²+b²

LHS=RHS

Hence proved.

Answer 2

Answer:

$\frac{a^2+\:b^2}{a^2-\:b^2}$

Step-by-step explanation:

$\red{\underline{Question:-}}$

$\bold{If \:tan\theta=\frac{a}{b},then\:\frac{a\:sin\:\theta+b\:cos\:\theta}{a\:sin\:\theta-b\:cos\:\theta}} =?$

$\red{\underline{Solution:-}}$

$\Rightarrow a\:sin \theta=\frac{a^2}{\sqrt{a^2+b^2}},b\:cos\theta=\frac{b^2}{\sqrt{a^2+b^2} }$

$\bold{Replacing\:values,} \frac{\frac{a^2}{\sqrt{a^2+b^2} }+\frac{b^2}{\sqrt{a^2+b^2} } }{\frac{a^2}{\sqrt{a^2+b^2} } -\frac{b^2}{\sqrt{a^2+b^2} } }$

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

                            $=\frac{a^2+b^2}{a^2-b^2}$

$\red{\boxed{\frac{a^2+b^2}{a^2-b^2}}} \:\bold{This,is\:Final\:answer}$