Question

If MsinA+NcosA=P & McosA-NsinA=Q then find M^2+N^2+P^2+Q^2

Answer 1

Answer:

$\Rightarrow 2(P^2+Q^2)$   is the required result.

Step-by-step explanation:

We have been given:

MsinA+NcosA=P     (1)

& McosA-NsinA=Q    (2)

We need to find $M^2+N^2+P^2+Q^2$

Squaring both sides in (1) and (2) and then adding them

After squaring both sides:

$(MsinA+NcosA)^2=P^2$

$M^2 sin^2A+N^2 cos^2A+2MNcosAsinA=P^2$     (3)

And $M^2 cos^2A+N^2sin^2A-2MNcosAsinA=Q^2$    (4)

Now, adding (3) and (4) we get:

$M^2 sin^2A+N^2 cos^2A+2MNcosAsinA+M^2 cos^2A+N^2sin^2A-2MNcosAsinA=P^2+Q^2$

After simplification:

$M^2sin^2A+N^2cos^2A+M^2cos^2A+N^2sin^2A=P^2+Q^2$

As we know: $Sin^2x+cos^2x=1$

$M^2sin^2A+M^2cos^2A+N^2cos^2A+N^2sin^2A=P^2+Q^2$

$M^2(sin^2A+cos^2A)+N^2(sin^2A+cos^2A)=P^2+Q^2$

$M^2+N^2=P^2+Q^2$

So, $M^2+N^2+P^2+Q^2$     (5)

Substitute $M^2+N^2=P^2+Q^2$   in (5) we get:

$P^2+Q^2+P^2+Q^2$

$\Rightarrow 2(P^2+Q^2)$   is the required result.