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Question
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Solution
`"tan A"/(1 + "tan"^2 "A")^2 + "cot A"/(1 + "cot"^2 "A")^2 = "sin A cos A"`.
LHS = `"tan A"/(1 + "tan"^2 "A")^2 + "cot A"/(1 + "cot"^2 "A")^2`
LHS = `"tan A"/("sec"^2 "A")^2 + "cot A"/("cosec"^2 "A")^2 ...{( 1 + "tan"^2θ = "sec"^2θ),(1 + "cot"^2θ = "cosec"^2θ):}`
LHS = `"tan A" × 1/("sec"^2 "A")^2 + "cot A" × 1/("cosec"^2 "A")^2`
LHS = `"sin A"/"cos A" × "cos"^4 "A" + "cos A"/"sin A" × "sin"^4 "A" ...{(cosθ = 1/sec θ),(sin θ = 1/"cosecθ"):}`
LHS = sinA cos3A + cosA sin3A
LHS = sinA cosA (cos2A + sin2A)
LHS = sinA cosA.(1) ...(cos2A + sin2A = 1)
LHS = sinA cosA
RHS = sinA cosA
LHS = RHS
Hence proved.
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