MCQ
$\cot 34^{\circ}-\tan 56^{\circ}=$ ?
- A$1$
- B$\frac{1}{2}$
- ✓$0$
- DNone of these
✓
Answer
Correct option: C.
$0$
$\cot 34^{\circ}-\tan 56^{\circ}$
$=\cot \left(90^{\circ}-56^{\circ}\right)-\tan 56^{\circ}$
$=\tan 56^{\circ}-\tan 56^{\circ}=0$
$\left[\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right]$
$=\cot \left(90^{\circ}-56^{\circ}\right)-\tan 56^{\circ}$
$=\tan 56^{\circ}-\tan 56^{\circ}=0$
$\left[\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right]$
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