InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If tanθ = a/b then the value of \(\cfrac{bsin\theta-acos\theta}{bsin\theta+acos\theta}\) is(1) 1(2) \(\frac{a^2-b^2}{a^2+b^2}\)(3) \(\frac{b^2-a^2}{b^2+a^2}\)(4) 0 |
|
Answer» (4) 0 \(\cfrac{bsin\theta-acos\theta}{bsin\theta+acos\theta}\) Dividing Nr. & Dr. by cosθ = \(\cfrac{btan\theta-a}{btan\theta+a}=\) \(\cfrac{b\times\frac{a}{b}-a}{b\times\frac{a}{b}+a}\) = \(\cfrac{0}{2a}\) = 0 |
|