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
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Solving,
⇒tan2A(sec A + 1)2⇒sec2A−1(sec A + 1)2⇒(sec A + 1)(sec A - 1)(sec A + 1)2⇒sec A - 1sec A + 1⇒1cos A−11cos A+1⇒1 - cos Acos A1 + cos Acos A⇒1 - cos A1 + cos A.\Rightarrow \dfrac{\text{tan}^2 A}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A - 1}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{(sec A + 1)(sec A - 1)}}{\text{(sec A + 1)}^2} \\[1em] \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}}{\text{1 + cos A}}.⇒(sec A + 1)2tan2A⇒(sec A + 1)2sec2A−1⇒(sec A + 1)2(sec A + 1)(sec A - 1)⇒sec A + 1sec A - 1⇒cos A1+1cos A1−1⇒cos A1 + cos Acos A1 - cos A⇒1 + cos A1 - cos A.
Hence, Option 2 is the correct option.
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cot4 θ + cot2 θ is equal to :
2 cot2 θ. cosec2 θ
tan2 θ + tan4 θ
tan2 θ.cosec2 θ
cosec4 θ - cosec2 θ
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
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
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
