Prove the following identities :
sec A - 1sec A + 1=1 - cos A1 + cos A\dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}}sec A + 1sec A - 1=1 + cos A1 - cos A
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Solving L.H.S. of the equation :
⇒sec A - 1sec A + 1⇒1cos A−11cos A+1⇒1 - cos Acos A1 + cos Acos A⇒(1 - cos A) × cos A(1 + cos A) × cos A⇒1 - cos A1 + cos A.\Rightarrow \dfrac{\text{sec A - 1}}{\text{sec A + 1}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos A}} - 1}{\dfrac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - cos A}}{\text{cos A}}}{\dfrac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A) × cos A}}{\text{(1 + cos A) × cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{1 + cos A}}.⇒sec A + 1sec A - 1⇒cos A1+1cos A1−1⇒cos A1 + cos Acos A1 - cos A⇒(1 + cos A) × cos A(1 - cos A) × cos A⇒1 + cos A1 - cos A.
Since, L.H.S. = R.H.S.
Hence, proved that sec A - 1sec A + 1=1 - cos A1 + cos A\dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}}sec A + 1sec A - 1=1 + cos A1 - cos A.
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11 - sin A\dfrac{1}{\text{1 - sin A}}1 - sin A1 is equal to :
1 + sin A
sec2 A (1 + sin A)
sec2 A
sec2 A + sin A
tan2A(sec A + 1)2\dfrac{\text{tan}^2 A}{\text{(sec A + 1)}^2}(sec A + 1)2tan2A is equal to :
1 + cos A1 - cos A\dfrac{\text{1 + cos A}}{\text{1 - cos A}}1 - cos A1 + cos A
1 - cos A1 + cos A\dfrac{\text{1 - cos A}}{\text{1 + cos A}}1 + cos A1 - cos A
11 + cos A\dfrac{1}{\text{1 + cos A}}1 + cos A1
11 - cos A\dfrac{1}{\text{1 - cos A}}1 - cos A1
1tan A + cot A=cos A sin A\dfrac{1}{\text{tan A + cot A}} = \text{cos A sin A}tan A + cot A1=cos A sin A
tan A - cot A = 1 - 2 cos2Asin A cos A\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}sin A cos A1 - 2 cos2A
