1.

The value of A that satisfies the equation a sin A + b cos A = c is equal to?1. \(\tan^{-1} \left(\dfrac{a}{b}\right) \pm \cos^{-1} \left(\dfrac{c}{\sqrt{a^2+b^2}}\right)\)2. \(\tan^{-1}\left(\dfrac{c}{b}\right)\pm \sin^{-1} \left(\dfrac{a}{\sqrt{a^2+b^2}}\right)\)3. \(\tan^{-1}\left(\dfrac{a}{b}\right)\pm \sin^{-1} \left(\dfrac{a}{\sqrt{a^2 +b^2}}\right)\)4. None

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


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