If a -b =c, then a +b = - Vidyadip

MCQ

If $\text{a}\cos\theta-\text{b}\sin\theta=\text{c},$ then $\text{a}\sin\theta+\text{b}\cos\theta=$

A
$\pm\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
✓
$\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
C
$\pm\sqrt{\text{c}^2-\text{a}^2-\text{b}^2}$
D
None of these.

✓

Answer

Correct option: B.

$\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$

Squaring,
$\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta-2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ \text{a}^2(1-\sin^2\theta)+\text{b}^2(1-\cos^2\theta)-2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ \text{a}^2-\text{a}^2\sin^2\theta+\text{b}^2(1-\cos^2\theta)-2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ -\text{a}^2\sin^2\theta-\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta=\text{c}^2-\text{a}^2-\text{b}^2$
$\Rightarrow\ \text{a}^2\sin^2\theta+\text{b}^2\cos\theta+2\text{ab}\sin\theta\cos\theta=\text{a}^2+\text{b}^2-\text{c}^2 $
$\ (\text{Dividing by}-1)$
$(\text{a}\sin\theta+\text{b}\cos\theta)^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\therefore\ \text{a}\sin\theta+\text{b}\cos\theta=\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$

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