Question

fast please with full explained

Answer 1

Answer:

tan⁻¹ (a/b) - x

Step-by-step explanation:

$Given:tan^{-1}[\frac{acosx-bsinx}{bcosx+asinx}]$

Divide by bcosx, we get

$=>tan^{-1}[\frac{\frac{acosx-bsinx}{bcosx}}{\frac{bcosx+asinx}{bcosx}}]$

$=>tan^{-1}[\frac{\frac{acosx}{bcosx}-\frac{bsinx}{bcosx}}{\frac{bcosx}{bcosx}+\frac{asinx}{bcosx}}]$

$=>tan^{-1}[\frac{\frac{a}{b}-tanx}{1+\frac{a}{b}tanx}]$

$=>tan^{-1}\frac{a}{b}-tan^{-1}(tanx)$

$=>\boxed{tan^{-1}\frac{a}{b} - x}}$

Hope it helps!

Answer 2

Step-by-step explanation:

The answer is explained below!