sin 5/x + sin 12/x = π÷2 in inverse
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Letsin−1x=θ⇒x=sinθ=cos(π2−θ)
⇒cos−1x=π2−θ=π2−sin−1x
∴sin−1x+cos−1x=π2
⇒cos−1x=π2−θ=π2−sin−1x
∴sin−1x+cos−1x=π2
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