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Evaluate..cot12°cot38°cot52°cot60°cot78°

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Evaluate..cot12°cot38°cot52°cot60°cot78°

Posted by Kabita Beura 7 years, 2 months ago

CBSE > Class 10 > Mathematics

4 answers

Himanshi Sharma 7 years, 2 months ago

1/root3

Ram Kushwah 7 years, 2 months ago

cot12cot38cot52cot60cot78

=cot(90-78)cot(90-52)cot52cot60cot78

=tan78cot78 x tan52cot52 x cot60

=1x1x1 x cot60

={tex}\frac{1}{\sqrt3}{/tex}

Vaishnavi Singh 7 years, 2 months ago

1/√3

Vishal Gorana 7 years, 2 months ago

1/√3

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