MCQ
In Fig. $\tan A-\cot C$ is equal to
- ✓$0$
- B$\frac{5}{12}$
- C$\frac{7}{13}$
- D$-\frac{7}{13}$
✓
Answer
Correct option: A.
$0$
(A)0
In right triangle $A B C$, we have $A B=12 cm, A C=13 cm$.
$
\begin{array}{ll}
\therefore & A C^2=A B^2+B C^2 \Rightarrow B C=\sqrt{A C^2-A B^2}=\sqrt{169-144}=5 \\
\therefore & \tan A=\frac{B C}{A B} \text { and } \cot C=\frac{B C}{A B} \\
\Rightarrow & \tan A=\frac{5}{12} \text { and } \cot C=\frac{5}{12} \Rightarrow \tan A-\cot C=0
\end{array}
$
In right triangle $A B C$, we have $A B=12 cm, A C=13 cm$.
$
\begin{array}{ll}
\therefore & A C^2=A B^2+B C^2 \Rightarrow B C=\sqrt{A C^2-A B^2}=\sqrt{169-144}=5 \\
\therefore & \tan A=\frac{B C}{A B} \text { and } \cot C=\frac{B C}{A B} \\
\Rightarrow & \tan A=\frac{5}{12} \text { and } \cot C=\frac{5}{12} \Rightarrow \tan A-\cot C=0
\end{array}
$
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