# TrigonometryProve q((p^2)-1) = 2

#### Wild ownz al

##### Member
If SinA + CosA = p and TanA + CotA = q, prove that q((p^2)-1) = 2. (Spent hours STILL could not figure out!)

##### Well-known member
If SinA + CosA = p and TanA + CotA = q, prove that q((p^2)-1) = 2. (Spent hours STILL could not figure out!)
Let us start from the LHS

$\sin\, A + \cos\,A = p$
square both sides
$(\sin\, A + \cos\,A)^2 = p^2$

or $\sin^2 A + 2 \sin\, A \cos\, A + cos^2 A = p^2$
or $1 +2 \sin\, A \cos\, A = p^2$
or $p^2 - 1 = 2 \sin\, A \cos\, A\cdots(1)$
from $2^{nd}$ condition
$\frac{\sin\, A}{\cos\, A} + \frac{\cos \, A}{\sin \, A} = q$
or $\frac{\sin^2 A+\cos^2 A}{\cos\, A\sin \, A} = q$
or $\frac{1}{\cos\, A\sin \, A} = q\cdots(2)$

multiplying (1) with (2) you get the result