(a) (i) Prove that : 2cos3θ−cosθsinθ−2sin3θ=tanθ
(ii) Prove that : sinA−cosAsinA+cosA+sinA+cosAsinA−cosA=2sin2A−12
OR
(b) If xsin3θ+ycos3θ=sinθcosθ and xsinθ=ycosθ, prove that x2+y2=1.
Get the complete, step-by-step math solution for: "(a) (i) Prove that : (sin θ - 2sin³ θ)/(2cos³ θ - cos θ) = tan θ (ii) Prove that : (sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) =...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Simplify the first expression
To prove the first identity, we start by factoring out common terms from the numerator and the denominator. In the numerator, we factor out sinθ, and in the denominator, we factor out cosθ.
Step 2: Apply trigonometric identities
We use the identity cos2θ=1−2sin2θ and cos2θ=2cos2θ−1. Substituting these into the expression, we see that the terms in the parentheses are identical.
Step 3: Simplify to tan θ
Since the terms (1−2sin2θ) and (2cos2θ−1) are both equal to cos2θ, they cancel out (assuming cos2θ=0). This leaves us with cosθsinθ, which is equal to tanθ.
Step 4: Combine fractions for the second expression
For the second identity, we combine the two fractions by finding a common denominator, which is (sinA−cosA)(sinA+cosA). We then cross-multiply the numerators.
Step 5: Expand and simplify the numerator
We expand the squared terms in the numerator. The terms 2sinAcosA and −2sinAcosA cancel each other out. In the denominator, we use the difference of squares formula, (a−b)(a+b)=a2−b2.
Step 6: Apply Pythagorean identity and simplify
Using the Pythagorean identity sin2A+cos2A=1 in the numerator, and substituting cos2A=1−sin2A in the denominator, we simplify the expression to the desired right-hand side.