Question

if tan theta equals to a /b prove that aSin theta minus b cos theta divided by asin theta + b cos theta equals to a square minus b square divided by a square + b square ​

Answer 1

Step-by-step explanation:

We have given that,

                                 Tan Θ = a / b;                          (i)

We asked to prove that,

                                 (a sinΘ - b cosΘ) / (a sin Θ + b cosΘ) = $\frac{ a^{2} - b^{2} }{ a^{2} + b^{2} }$.     (ii)

So, divide the numerator and the denominator by cos Θ of the LHS of equation (ii);

                               ⇒ {(a sinΘ - b cosΘ) / cosΘ } / {(a sin Θ + b cosΘ) / cosΘ}

∴As we know that when divide any number both in the numerator and denominator on a fraction then its value will be equal.

Hence, this will give us,

                              ⇒  {a tan Θ - b}  /  {a tan Θ + b};      (iii)

Now, put the value of Tan Θ in equation (iii) from equation (ii),

                               ⇒{a * a/b - b}  /  {a * a/b  +  b};

                               ⇒{$a^{2}$ / b  -  b}  /   {$a^{2}$ / b  + b} ;

                               ⇒ {($a^{2}$ - $b^{2}$) / b }  /  {($a^{2}$ + $b^{2}$) / b }  

From here b will cancel out,

Therefore,                 ⇒ ($a^{2}$ - $b^{2}$)   /  ($a^{2}$ + $b^{2}$);

                             ⇒ $\frac{ a^{2} - b^{2} }{ a^{2} + b^{2} }$.

Hence, proved.

That's all

Answer 2

Refers to attachment ☆~