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The expression 2 cos2α – 1 can be simplified to

व्यंजक 2 cos2α – 1 का सरलीकरण रूप हो सकता है

1 – 4 sin²α

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4 cos²α – 3

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\frac{cos 3 α}{cos α}

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3 – 4 sin²α

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Solution

The correct option is C $\frac{cos 3 α}{cos α}$
Solution

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Similar questions

The correct statement with regard to $H_{2}^{+}$ & $H_{2}^{-}$ is

H_{2} + B r_{2} \to 2 H B r

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}] \frac{d [H_{2} O]}{d t} = k [N O] [H_{2}] - \frac{d [N O]}{d t} = k_{1}^{'} [N O] [H_{2}] - \frac{d [H_{2}]}{d t} = k_{1}^{''} [N O] [H_{2}]

k, k_{1}, k_{1}^{'}

k_{1}^{''}

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

\frac{d [N_{2}]}{d t} = k_{1} [N O] [H_{2}] \frac{d [H_{2} O]}{d t} = k [N O] [H_{2}] - \frac{d [N O]}{d t} = k_{1}^{'} [N O] [H_{2}] - \frac{d [H_{2}]}{d t} = k_{1}^{''} [N O] [H_{2}]

k, k_{1}, k_{1}^{'}

k_{1}^{''}

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

{d t [N_{2}] / d t} = k_{1} [N O] [H_{2}]; {d [H_{2} O] / d t) = k [N O] [H_{2}];

{- d [N O] / d t} = k_{1}^{'} [N O] [H_{2}]; {- d [H_{2}] / d t} = k_{1}^{''} [N O] [H_{2}]

k, k_{1}, k_{1}^{'}

k_{1}^{''}

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