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Show that tan 48° tan 16° tan 42° tan 74° = 1 |
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Answer» L.H.S. = tan 48° tan 16° tan 42° tan 74° = tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°) = tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ] = tan 48° . tan 16° . \(\frac{1}{tan \,48^∘}\) . \(\frac{1}{tan \,16^∘}\) [∵ cot θ = \(\frac{1}{tan \,θ}\)] = 1 = R.H.S. ∴ L.H.S. = R.H.S. |
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