Question

Pls tell the answer fast we are waiting​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

if $\tan\theta=\frac{a}{b}$

Prove that,

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

LHS

$\frac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}\\\\\Rightarrow \frac{\frac{a\sin\theta-b\cos\theta}{\cos\theta}}{\frac{a\sin\theta+b\cos\theta}{\cos\theta}}\text{ [dividing both numerator and denominator by }\cos\theta]\\\\\Rightarrow \frac{\frac{a\sin\theta}{\cos\theta}-\frac{b\cos\theta}{\cos\theta}}{\frac{a\sin\theta}{\cos\theta}+\frac{b\cos\theta}{\cos\theta}}\\\\\Rightarrow\frac{a\tan\theta-b}{a\tan\theta+b}$

$\\\\\Rightarrow\frac{a\times\frac{a}{b}-b}{a\times\frac{a}{b}+b}\text{ [putt the value of }\tan\theta=\frac{a}{b}]\\\\\Rightarrow\frac{\frac{a^2}{b}-b}{\frac{a^2}{b}+b}\\\\\Rightarrow\frac{\frac{a^2-b^2}{b}}{\frac{a^2+b^2}{b}}\\\\\Rightarrow\frac{a^2-b^2}{a^2+b^2}$

therefore,

LHS=RHS (Proved)