Solve tan20tan40tan80
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this is the answer
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let 20=A
tan A tan(60-A)tan(60+A)
=sin A sin(60-A) sin(60+A) / cos A cos(60-A) cos(60+A)
=sinA (sin²60-sin²A) / cosA (cos²A-sin²60)
=sinA (3/4 - sin²A) /cosA ( cos²A - 3/4 )
= 1/4* sinA ( 3- 4 sin²A) / 1/4 cosA ( 4 cos²A - 3)
= 1/4 * ( 3sinA - 4 sin³A) / 1/4 ( 4cos³A -3 cosA)
= 1/4 sin 3A /1/4 cos3A
= sin3A / cos 3A
= tan 3A
= tan (3* 20 )
= tan ( 60)
= √3
tan A tan(60-A)tan(60+A)
=sin A sin(60-A) sin(60+A) / cos A cos(60-A) cos(60+A)
=sinA (sin²60-sin²A) / cosA (cos²A-sin²60)
=sinA (3/4 - sin²A) /cosA ( cos²A - 3/4 )
= 1/4* sinA ( 3- 4 sin²A) / 1/4 cosA ( 4 cos²A - 3)
= 1/4 * ( 3sinA - 4 sin³A) / 1/4 ( 4cos³A -3 cosA)
= 1/4 sin 3A /1/4 cos3A
= sin3A / cos 3A
= tan 3A
= tan (3* 20 )
= tan ( 60)
= √3
sukhetha:
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