Question

sinA+cosA=P then sinA-cosA=?

Answer 1

square the first equation we get sin^2A+cos^2A+2sinAcosA=p^2 from this we get 2sinAcosA=p^2 -1 take it as equation 1 then square the given second equation we get sin^2A+cos^2A-2sinAcosA substitute equation 1 in 2 we get 1-(p^2 -1) square root the value then we get √(2-p^2)

Answer 2

Given:

$\sin A$ + $\cos A$ = P             .......... (1)

Let $\sin A$ - $\cos A$ = x        .......... (2)

We have to find, the value of $\sin A$ - $\cos A$ = ?

Solution:

Squaring and adding equations (1) and (2), we get

$(\sin A+\cos A)^2$ + $(\sin A-\cos A)^2$ = $P^2 + x^2$

⇒ $P^2 + x^2$ = $\sin^2 A+\cos^2 A+2\sin A\cos A$ + $\sin^2 A+\cos^2 A-2\sin A\cos A$

⇒ $P^2 + x^2$ = $\sin^2 A+\cos^2 A$ + $\sin^2 A+\cos^2 A$

⇒ $P^2 + x^2$ = $2\sin^2 A+2\cos^2 A$

⇒ $P^2 + x^2$ = $2(\sin^2 A+\cos^2 A)$

Using the trigonometric identity:

$\sin^2 A$ + $\cos^2 A$ = 1

⇒ $P^2 + x^2$ = 2(1)

⇒ $P^2 + x^2$ = 2

⇒ $x^2$ = 2 - $P^2$

⇒ x = $\sqrt{2-P^2}$

∴ $\sin A$ - $\cos A$ = $\sqrt{2-P^2}$

Thus, if $\sin A$ + $\cos A$ = P, then $\sin A$ - $\cos A$ = $\sqrt{2-P^2}$ .