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What is the value of [sin(x + y) - 2sinx + sin(x - y)]/[cos(x - y) + cos(x + y) - 2cosx] × [(sin 10x - sin 8x)/(cos 10x + cos 8x)]?
- A0
- Btan2x
- C1
- D2tanx
Answer not set
Discussion 1
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CCommunity
Simplifying the given equation by applying the formula:
sinC + sinD = 2[sin(C + D)/2.cos(C - D)/2]
cosC + cosD = 2[cos(C + D)/2.cos(C - D)/2]
$(\Rightarrow \frac{{2\sin x\cos y\; - \;2\sin x}}{{2\cos x\cos y\; - \;2\cos x}} \times \;\frac{{2\sin x\cos 9x}}{{2\cos x\;cox\;9x}})$
$(\Rightarrow \frac{{\sin x(\cos y\; - \;1)}}{{\cos x(\cos y\; - \;1)}} \times \;\frac{{\sin x}}{{\cos x\;}})$ = tan2x
∴ [sin(x + y) - 2sinx + sin(x - y)]/[cos(x - y) + cos(x + y) - 2cosx] × [(sin 10x - sin 8x)/(cos 10x + cos 8x)] = tan2x
sinC + sinD = 2[sin(C + D)/2.cos(C - D)/2]
cosC + cosD = 2[cos(C + D)/2.cos(C - D)/2]
$(\Rightarrow \frac{{2\sin x\cos y\; - \;2\sin x}}{{2\cos x\cos y\; - \;2\cos x}} \times \;\frac{{2\sin x\cos 9x}}{{2\cos x\;cox\;9x}})$
$(\Rightarrow \frac{{\sin x(\cos y\; - \;1)}}{{\cos x(\cos y\; - \;1)}} \times \;\frac{{\sin x}}{{\cos x\;}})$ = tan2x
∴ [sin(x + y) - 2sinx + sin(x - y)]/[cos(x - y) + cos(x + y) - 2cosx] × [(sin 10x - sin 8x)/(cos 10x + cos 8x)] = tan2x