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Given that tan A= 5/12 and tan B= -4/3 such that A is an acute angle and B is an obtuse angle find the value of,a) tan (A-45°)b) tan (B+360°)

Given That Tan A 512 And Tan B 43 Such That A Is An Acute Angle And B Is An Obtuse Angle Find The Value Ofa Tan A45b Tan B360 class=

Sagot :

Solve for tan(A-45°).

Recall that tan(45°) = 1

[tex]\begin{gathered} \tan (A-45\degree)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ \tan (A-45\degree)=\frac{\frac{5}{12}-1}{1+(\frac{5}{12})(1)} \\ \tan (A-45\degree)=\frac{-\frac{7}{12}}{1+\frac{5}{12}} \\ \tan (A-45\degree)=\frac{-\frac{7}{12}}{\frac{17}{12}} \\ \tan (A-45\degree)=-\frac{7}{17} \end{gathered}[/tex]

Therefore, tan(A-45°) = -7/17.

Solve for tan(B+360°)

Recall that tan(360°) = 0

[tex]\begin{gathered} \tan (B+360\degree)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \tan (B+360\degree)=\frac{\frac{5}{12}+0}{1-(\frac{5}{12})(0)} \\ \tan (B+360\degree)=\frac{\frac{5}{12}}{1-\frac{5}{12}} \\ \tan (B+360\degree)=\frac{\frac{5}{12}}{\frac{7}{12}} \\ \tan (B+360\degree)=\frac{5}{7} \end{gathered}[/tex]

Therefore, tan(B+360°) = 5/7.

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