MCQ
The value of ${\left(17 \sec ^2 29^{\circ}-17 \cot ^2 61^{\circ}\right)} $ is equal to :
- A$34$
- B$0$
- ✓$17$
- D$1$
✓
Answer
Correct option: C.
$17$
To find the value of ${\left[17 \sec ^2 29^{\circ}-17 \cot ^2 61^{\circ}\right]} $
Now,
${\left[17 \sec ^2 29^{\circ}-17 \cot ^2 61^{\circ}\right]} $
$ =\left[17\left(1+\tan ^2 29^{\circ}\right)-17 \cot ^2 61^{\circ}\right] $
$ =\left[17+17 \tan ^2 29^{\circ}-17 \cot ^2 61^{\circ}\right] $
$ =\left[17+17 \tan ^2\left(90^{\circ}-61^{\circ}\right)-17 \cot ^2 61^{\circ}\right] $
$ =\left[17+17 \cot ^2 61^{\circ}-17 \cot ^2 61^{\circ}\right] $
$= 17$
Now,
${\left[17 \sec ^2 29^{\circ}-17 \cot ^2 61^{\circ}\right]} $
$ =\left[17\left(1+\tan ^2 29^{\circ}\right)-17 \cot ^2 61^{\circ}\right] $
$ =\left[17+17 \tan ^2 29^{\circ}-17 \cot ^2 61^{\circ}\right] $
$ =\left[17+17 \tan ^2\left(90^{\circ}-61^{\circ}\right)-17 \cot ^2 61^{\circ}\right] $
$ =\left[17+17 \cot ^2 61^{\circ}-17 \cot ^2 61^{\circ}\right] $
$= 17$
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