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Question Number 187375 by Rupesh123 last updated on 16/Feb/23

Commented by mokys last updated on 16/Feb/23

$$\frac{\frac{cosx-sinx}{cosx}}{\frac{sinx-cosx}{sinx}}=\frac{-secx}{cscx}=-tanx\ne 2sinx$$

Commented by Frix last updated on 17/Feb/23

$$\mathrm{Wrong}.$$$$-\tan x=2\sin x$$$$0⩽x<2\pi$$$$x\in \{0,\frac{2\pi }{3},\pi ,\frac{4\pi }{3}\}$$

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