Question

if tan theta is equal to a/b ,find Asin theta -bcos theta/ Asin theta+bcostheta

Answer 1

Let (theta) be (x)

tan(x) = a/b

Now,

a sin(x) - b cos(x)
_____________
a sin(x) + b cos(x)

Dividing numerator and denominator by cos(x),

a tan(x) - b
= ________
a tan(x) + b

a (a/b) - b
= _______
a(a/b) + b

= (a^2 - b^2) / (a^2 + b^2)

Hope this helps... plsss make it the brainliest

Answer 2

Given,
tan$\theta=\dfrac{a}{b}$


From the properties of trigonometric ratios,
we know : -

tan$\theta=\dfrac{sin\theta}{cos\theta}$


Now,
Comparing the values of tan$\theta$ from the trigonometric functions and question



Therefore,
$\implies \dfrac{sin \theta}{cos \theta} = \dfrac{a}{b} \\ \\ \\ \implies \dfrac{a}{b} \times \dfrac{sin \theta}{cos \theta} = \dfrac{a}{b} \times \dfrac{a}{b} \\ \\ \\ \implies \dfrac{a \: sin \theta}{b \: cos \theta} = \dfrac{a {}^{2} }{ {b}^{2} }$


From the properties of ratio and proportion, we know : -
If a : b = x : y
So, ( a + b ) : ( a - b ) = ( x + y ) : ( x - y )

This property is known as "Componendo and dividendo".



On the basis of the property given above : -

$\implies \dfrac{a \: sin \theta + b \: cos \theta}{a \: sin \theta - b \: cos \theta} = \dfrac{a {}^{2} + b {}^{2} }{a {}^{2} - b {}^{2} } \\ \\ \\ \implies \dfrac{a {}^{2} - b {}^{2} }{a {}^{2} + b {}^{2} } =\dfrac{a \: sin \theta - b \: cos \theta}{a \: sin \theta +b \: cos \theta}$



Hence,
value of $\dfrac{a \: sin \theta - b \: cos \theta}{a \: sin \theta +b \: cos \theta}\: is\: \dfrac{a {}^{2} - b {}^{2} }{a {}^{2} + b {}^{2} }$