sec a + tan a = x then find the value of sec a
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Explanation:
[tex] \tt We\:know\:the\: identity \\ \\ \tt 1 + {tan}^{2}A = {sec}^{2}A \\ \\ \tt \longrightarrow {sec}^{2}A - {tan}^{2}A = 1 \\ \\ \tt \longrightarrow ( secA + tanA )( secA - tanA ) = 1 \\ \\ \tt \longrightarrow secA - tanA = \frac{1}{secA + tanA } \\ \\ \tt So \: if \:secA + tanA = x \: then \: secA - tanA = \frac{1}{x} \\ \\ \tt Adding\:above\:two\:equations \\ \\ \tt \longrightarrow 2secA = x + \frac{1}{x} \\ \\ \tt \longrightarrow 2secA = \frac{{x}^{2}+1}{x} \\ \\ \tt \longrightarrow \boxed{\bold{\red{\tt secA = \frac{{x}^{2}+1}{2x} }}} [/tex]
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Answer:
Sec a = x² + 1/2x.
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