Correct Answer – Option 1 : \(\tan^{-1} \left(\dfrac{a}{b}\right) \pm \cos^{-1} \left(\dfrac{c}{\sqrt{a^2+b^2}}\right)\)

Calculation:

Given: a sin A + b cos A = c

Divide both sides by \(\rm \frac {1}{\sqrt {a^2 +b^2}}\), we get

⇒ \(\rm \frac {a}{\sqrt {a^2 +b^2}}\) sin A + \(\rm \frac {b}{\sqrt {a^2 +b^2}}\) cos A = \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

Let sin α = \(\rm \frac {a}{\sqrt {a^2 +b^2}}\) and cos α = \(\rm \frac {b}{\sqrt {a^2 +b^2}}\)

⇒ sin A sin α + cos A cos α = \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

⇒ cos (A – α) = \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

⇒ A – α = cos^-1 \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

⇒ A = scos-1 \(\rm \frac {c}{\sqrt {a^2 +b^2}}\) + α

Now, \(\rm \tan α = \frac {\sin α}{\cos α} = \frac ab\)

∴ α = tan^-1 \(\rm \frac ab\)

So, A = cos-1 \(\rm \frac {c}{\sqrt {a^2 +b^2}}\) + tan-1 \(\rm \frac ab\) = \(\rm \tan^{-1} \left(\frac{a}{b}\right) + \cos^{-1} \left(\frac{c}{\sqrt{a^2+b^2}}\right)\)