secA-tanA=x then secA+tanA?
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Heya mate ❤️
▪️Given secA + tanA = x → (1) Recall that sec2A − tan2A = 1 ⇒ (secA + tanA)(secA − tanA) = 1 ⇒x (secA − tanA) = 1 ⇒(secA − tanA) = 1/x → (2) Subtract (2) from (1), we get (secA + tanA) − (secA − tanA) = x − (1/x) ⇒ secA + tanA − secA + tanA = (x2 − 1)/x ⇒ 2tanA = (x2 − 1)/x ∴ tanA = (x2 − 1)/ 2x❤️
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▪️Given secA + tanA = x → (1) Recall that sec2A − tan2A = 1 ⇒ (secA + tanA)(secA − tanA) = 1 ⇒x (secA − tanA) = 1 ⇒(secA − tanA) = 1/x → (2) Subtract (2) from (1), we get (secA + tanA) − (secA − tanA) = x − (1/x) ⇒ secA + tanA − secA + tanA = (x2 − 1)/x ⇒ 2tanA = (x2 − 1)/x ∴ tanA = (x2 − 1)/ 2x❤️
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