Trigonometry Question

**raj1985** · February 10th 2009, 10:15 PM

Please Help Me in proving this----- If 3secA-7tanA=3, prove that 3tanA-7secA=7

**Soroban** · February 11th 2009, 12:05 PM

Hello, raj1985!

Is there a typo? .The statement isn't true.

Prove: .If $3\sec A-7\tan A\:=\:3$ , then: . $3\tan A-7\sec A\:=\:{\color{red}-}7$ . ??

We have: . $3\sec A \:=\:3 + 7\tan A$

Square both sides: . $9\sec^2\!A \:=\:9 + 42\tan A + 49\tan^2\!A$

. . . . . . . . . . . $9(\tan^2\!A + 1) \:=\:9 + 42\tan A + 49\tan^2\!A$

. . . . . . . . . . . $9\tan^2\!A + 9 \:=\:9 + 42\tan A + 49\tan^2\!A$

. . . . . . $40\tan^2\!A - 42\tan A \:=\:0$

. . . . . . $2\tan\theta(20\tan A - 21) \:=\:0$

We have: . $2\tan\theta \:=\:0\quad\Rightarrow\quad \theta \,=\,0$ . for which the statement is true.

And: . $20\tan A - 21 \:=\:0\quad\Rightarrow\quad \tan A \:=\:\frac{21}{20}$

. . . $\sec^2\!A \:=\:\tan^2\!A + 1 \:=\:\left(\frac{21}{20}\right)^2 + 1 \:=\:\frac{841}{400} \quad\Rightarrow\quad \sec A \:=\:\frac{29}{20}$

Then: . $3\tan A - 7\sec A \;=\;3\left(\frac{21}{20}\right) - 7\left(\frac{29}{20}\right) \;=\;\frac{63}{20} - \frac{203}{20} \;=\;\frac{\text{-}140}{20} \;=\;{\color{red}-7}$

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