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Mathematics

Prove that :

cot A - 12 - sec2A=cot A1 + tan A\dfrac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}

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Answer

Solving L.H.S. of the above equation :

⇒1tan A−12 - (1 + tan2A)⇒1 - tan Atan A2−1−tan2A⇒1 - tan Atan A(1 - tan2A)⇒1 - tan Atan A(1 - tan A)(1 + tan A)⇒1tan A(1 + tan A)⇒11cot A(1 + tan A)⇒cot A1 + tan A.\Rightarrow \dfrac{\dfrac{1}{\text{tan A}} - 1}{\text{2 - (1 + tan}^2 A)} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - tan A}}{\text{tan A}}}{2 - 1 - \text{tan}^2 A} \\[1em] \Rightarrow \dfrac{\text{1 - tan A}}{\text{tan A(1 - tan}^2 A)} \\[1em] \Rightarrow \dfrac{\text{1 - tan A}}{\text{tan A(1 - tan A)(1 + tan A)}} \\[1em] \Rightarrow \dfrac{1}{\text{tan A(1 + tan A)}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{\text{cot A}} \text{(1 + tan A)}} \\[1em] \Rightarrow \dfrac{\text{cot A}}{\text{1 + tan A}}.

Since, L.H.S. = R.H.S.

Hence, proved that cot A - 12 - sec2A=cot A1 + tan A\dfrac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}.

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