Prove that : = sec A cosec A + 1

By solving L.H.S. of the equation : By formula, sin 2 A + cos 2 A = 1 Since, L.H.S. = R.H.S. Hence, proved that = sec A cosec A + 1.

Mathematics

Prove that :
$\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1

Trigonometric Identities

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Answer

By solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}} \\[1em] \Rightarrow \dfrac{\text{tan A}}{1 - \dfrac{1}{\text{tan A}}} + \dfrac{\dfrac{1}{\text{tan A}}}{\text{1 - tan A}} \\[1em] \Rightarrow \dfrac{\text{tan A}}{\dfrac{\text{tan A - 1}}{\text{tan A}}} + \dfrac{1}{\text{tan A(1 - tan A)}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A - 1}} + \dfrac{1}{\text{tan A(1 - tan A)}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A - 1}} - \dfrac{1}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \dfrac{\text{tan}^3 A - 1}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \dfrac{\text{(tan A - 1)(tan}^2 A + \text{ tan A + 1)}}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A + \text{ tan A + 1}}{\text{tan A}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A}} + \dfrac{\text{tan A}}{\text{tan A}} + \dfrac{1}{\text{tan A}} \\[1em] \Rightarrow \text{tan A + 1 + cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + 1 + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{sin A cos A + cos}^2 A}{\text{sin A cos A}}.$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{1 + \text{sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \Rightarrow \dfrac{1}{\text{sin A cos A}} + \dfrac{\text{sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \text{cosec A sec A} + 1.$

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

Hence, proved that $\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1.

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Prove that :
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Trigonometric Identities

Answer

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