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Mathematics

If sec A + tan A = p, show that :

sin A = p2−1p2+1\dfrac{p^2 - 1}{p^2 + 1}

Trigonometric Identities

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Answer

Substituting value of p in R.H.S. of the above equation :

⇒(sec A + tan A)2−1(sec A + tan A)2+1⇒sec2A+tan2A+2 sec A tan A−1sec2A+tan2A+2 sec A tan A+1\Rightarrow \dfrac{\text{(sec A + tan A)}^2 - 1}{\text{(sec A + tan A)}^2 + 1} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A + \text{tan}^2 A + \text{2 sec A tan A} - 1}{\text{sec}^2 A + \text{tan}^2 A + \text{2 sec A tan A} + 1}

By formula,

sec2 A - 1 = tan2 A and sec2 A = 1 + tan2 A

⇒sec2A−1+tan2A+2 sec A tan Asec2A+tan2A+1+2 sec A tan A⇒tan2A+tan2A+2 sec A tan Asec2A+sec2A+2 sec A tan A⇒2 tan2A+2 sec A tan A2 sec2A+2 sec A tan A⇒2tan A(tan A + sec A)2 sec A(sec A + tan A)⇒tan Asec A⇒sin Acos A1cos A⇒sin Acos A×cos A⇒sin A.\Rightarrow \dfrac{\text{sec}^2 A - 1 + \text{tan}^2 A + \text{2 sec A tan A}}{\text{sec}^2 A + \text{tan}^2 A + 1 + \text{2 sec A tan A}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A + \text{tan}^2 A + \text{2 sec A tan A}}{\text{sec}^2 A + \text{sec}^2 A + \text{2 sec A tan A}} \\[1em] \Rightarrow \dfrac{\text{2 tan}^2 A + \text{2 sec A tan A}}{\text{2 sec}^2 A + \text{2 sec A tan A}} \\[1em] \Rightarrow \dfrac{2\text{tan A}(\text{tan A + sec A})}{\text{2 sec A(sec A + tan A)}} \\[1em] \Rightarrow \dfrac{\text{tan A}}{\text{sec A}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} \times \text{cos A} \\[1em] \Rightarrow \text{sin A}.

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

Hence, proved that sin A = p2−1p2+1\dfrac{p^2 - 1}{p^2 + 1}.

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