Question

If $tan\Theta=\frac{a}{b}$, then a $\frac{asin\Theta+bcos\Theta}{asin\Theta-bcos\Theta}$ is equal to
(a)$\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$
(b)$\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
(c)$\frac{a+b}{a-b}$
(d)$\frac{a-b}{a+b}$

Answer 1

SOLUTION :  

The correct option is (a) : (a² + b²) / (a² - b²).

Given : tan θ = a/b

In right angle ∆ ,  

tan θ =  perpendicular/base = a/b  

perpendicular = a , base = b

Hypotenuse = √( perpendicular)² + (Base)²

[By Pythagoras theorem]

Hypotenuse = √ a² + b² = √a² + b²  

Hypotenuse = √a² + b²

sinθ  = perpendicular/hypotenuse = a/ √a² + b²

cos θ = base/ hypotenuse = b/ √a² + b²

The value of (a sin θ + b cos θ  ) / (a sin θ -  b cos θ )  :  

= [a(a/ √a² + b²) + b(b/ √a² + b²)] /  [a(a/ √a² + b²) - b(b/ √a² + b²)]

= (a² + b² / a² + b² ) / (a² - b² / a² + b²)

=  (a² + b² / a² + b² ) × (a² + b²) / (a² - b²)

(a sin θ + b cos θ  ) / (a sin θ -  b cos θ ) = (a² + b²) / (a² - b²)

Hence, the value of (a sin θ + b cos θ  ) / (a sin θ -  b cos θ ) is (a² + b²) / (a² - b²).

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Answer 2

The correct option is a.