If \(sin \ β + cos \ β = u\) and \(sec \ β + cosec \ β = v\), then show that \(v(u^2 - 1) = 2u\)
Proof:
\(sin \ β + cos \ β = u\)
Squaring on both sides, we get:
\((sin \ β + cos \ β)^2 = u^2\)
\(sin^2 \ β + cos^2 \ β+ 2 \ sin \ β \ cos \ β = u^2\)
Using the trigonometric identity, \(sin^2 \ β + cos^2 β=\) in the above equation, we have, \(u^2=\) ---(1)
Consider \(sec \ β + cosec \ β = v\)
\(\frac{u}{v}=\) ---(2)
Equating (1) and (2) then simplifying we get, \(2u = v(u^2 - 1)\)
Answer variants:
\(1 + 2 \ sin \ β \ cos \ β\)
\(1\)
\(2 + 2 \ sin \ β \ cos \ β\)
\(2\)
\(sin \ β \ cos \ β\)
\(0\)