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Question
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
Solution
We have,
LHS = `(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
= `cos(90° - 20°)/(sin 20°) + cos(90° - 31°)/(sin 31°) - 8 xx (1/2)^2`
= `(sin 20°)/(sin 20°) + (sin 31°) /(sin 31°) - 8 xx 1/4`
= 1 + 1 - 2
= 2 -2
= 0
= RHS
Hence proved.
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