Question

Without using trigonometric tables , a 10th grade student found out that tan7*tan23*tan60*tan67*tan83* = √3. Do you agree with this result ? if yes then justify your answer.

Answer 1

Using
tanA = cot(90-A),
$\: \: \: \: \tan(7) \times \tan(23) \times \tan(60) \times \tan(67) \times \tan(83) \\ = \tan(90 - 83) \times \tan(90 - 67) \times \tan(60) \times\tan(23) \times \tan(83) \\ = \cot(83) \times \cot(67)\tan(60) \times\times \tan(67) \times \tan(83) \\ = \cot(83) \times \tan(83) \times\tan(60) \times \cot(67) \times \tan(67) \\ = 1 \times \sqrt{3} \\ = \sqrt{3} .$

So the 10th student founded right answer.


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Answer 2

Answer:

Yes, he is right.

Step-by-step explanation:

tan 7 x tan 23 x tan 60 x tan 67 x tan 83

= tan 7 x cot(90-7) x tan 23 x tan 60 x cot (90-23) ; [cot (90 - A) = cot A]

= tan 7 x cot 7 x tan 23 x cot 23 x tan 60

= tan 60

= _/3