If a +b =4 and a -b =3, then a^2 + b^2 = 0

MCQ

If $\text{a}\cos\theta+\text{b}\sin\theta=4\text{ and a}\sin\theta-\text{b}\cos\theta=3,$ then $a^2 + b^2 = 0$

A
$7$
B
$12$
✓
$25$
D
None of these.

✓

Answer

Correct option: C.

$25$

$25$
Given,
$\text{a}\cos\theta+\text{b}\sin\theta=4,$
$\text{a}\sin\theta-\text{b}\cos\theta=3$
Squaring and then adding the above two equations, we have
$(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2=(4)^2+(3)^2$
$\Rightarrow\ (\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta)+2\times\text{a}\cos\theta\times\sin\theta)$
$=(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{a}\sin\theta\times\text{b}\cos\theta)=16+9$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta$
$=\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta=25$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta=25$
$\Rightarrow\ ( \text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta)+(\text{b}^2\sin^2\theta+\text{b}^2\cos^2\theta)=25$
$\Rightarrow\ \text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)=25$
$\Rightarrow\ \text{a}^2(1)+\text{b}^2(1)=25$
$\Rightarrow\ \text{a}^2+\text{b}^2=25$
Hence, the correct option is (C).