Prove: $\dfraf{cos³A + sin³A}{cosA + sinA} = \frac{1}{2} (2 - sin2A)$ Solution: LHS $= \dfrac{cos³A + sin³A}{cosA + sinA}$ $= \dfrac{(cosA)^3 + (sinA)^3}{cosA + sinA}$ Expand the numerator using the factorisation formula of (a³+b³) $= \dfrac{(cosA + sinA)(cos²A - sinAcosA + sin²A)}{(cosA + sinA)}$ $= cos²A - sinAcosA + sin²A$ $= (co…

Prove the following identity: $\dfrac{sin4A}{cos2A} × \dfrac{1 - cos2A}{1 - cos4A} = tanA$ Solution: LHS $= \dfrac{sin4A}{cos2A} × \dfrac{1 - cos2A}{1 - cos4A}$ $= \dfrac{2sin2Acos2A}{cos2A} × \dfrac{1 - cos2A}{1 - cos4A}$ $= 2sin2A × \dfrac{1 - cos2A}{1 - cos4A}$ Remember this identity: (1 - cos2A) = 2sin²A $= 2sin2A × \dfrac{2sin²…

Prove that: $\dfrac{1 + cos2A}{sin2A} = cotA$. Solution: LHS $= \dfrac{1 + cos2A}{sin2A}$ Use Trigonometry Identity of 1 and multiple angle formula of cos2A, we get, $= \dfrac{(sin²A+cos²A)+(cos²A - sin²A)}{sin2A}$ $= \dfrac{sin²A + cos²A + cos²A - sin²A}{sin2A}$ $= \dfrac{2cos²A}{sin2A}$ $= \dfrac{2cos²A}{2sinAcosA}$ $= \dfrac{2×co…

Question: Prove that: $\dfrac{cos2A}{1+ sin2A} = \dfrac{1- tanA}{1+ tanA}$ Solution: Taking LHS, $= \dfrac{cos2A}{1+ sin2A}$ Expanding cos2A and sin2A and identity of 1 $= \dfrac{cos^2A - sin^2A}{sin^2A + cos^2A + sin2A}$ Performing simple mathematical operations $= \dfrac{(cosA + sinA)(cosA - sinA)}{(cosA + sinA)^2}$ $= \dfrac{cosA…

Prove that: $\dfrac{1 - sin2A}{cos2A} = sec2A - tan2A$. Solution: Taking LHS $= \dfrac{1 - sin2A}{cos2A}$ Separately write the terms now $= \dfrac{1}{cos2A} - \dfrac{sin2A}{cos2A}$ Use the trigonometric identities now $= sec2A - tan2A$ = RHS #proved

Question: Prove the following trigonometric identity: $1 + tan 4A.tan 2A = sec 4A$ Solution: Taking LHS $ = 1 + tan 4A.tan 2A$ $= 1 + tan(2×2A) . tan2A$ Remember: $tan2A = \dfrac{2 tanA}{1 - tan^2 A}$ $= 1 + \dfrac{2tan2A}{1 - tan^2 2A} . tan2A$ $= 1 + \dfrac{2 tan^2 2A}{1 - tan^2 2A}$ $= \dfrac{(1 - tan^2 2A) + 2tan^2 2A}{1 - tan^2…