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# The value of tan81 - tan63 - tan27 + tan9 is......(angles are in degrees)Pls explain.....

manu saxena
46 Points
9 years ago

tan 81 -tan63 +tan 9 - tan 27=

tan(90-9)-tan(90-27)+tan9-tan27

=cot9-cot27+tan9-tan27

=(sec^2(9)/tan(9))-(sec^2(27)/tan(27))

=(1/(sin(9)cos(9))-((1/(sin(27)cos(27))

=(2/sin(18))-(2/(sin(54))

=4

Hardik singh
19 Points
3 years ago
Q=tan 9 - tan 27 - tan 63 + tan 81 = tan 9 - tan 27 - tan (90-27) + tan (90-9) = tan 9 - tan 27 - cot 27 + cot 9 =[tan 9 + cot 9] - [ tan 27 + cot 27] =[sin 9/ cos 9 + cos 9/ sin 9] - [ sin 27/cos 27 + cos 27/sin 27] = 2/sin 18 - 2/sin 54 = 2[ 1/sin 18 - 1/ sin 54] =2 [ (sin 54 - sin 18)/ ( sin 18 sin 54) ] = 2 [ ( 2 cos 36 sin 18)/( sin 18 sin 54)] = 4 cos 36 / sin 54 = 4 cos(90 - 54) / sin 54 = 4 sin 54 / sin 54 = 4 Ans
Azeem ali
15 Points
2 years ago

tan 81 -tan63 +tan 9 - tan 27=

tan(90-9)-tan(90-27)+tan9-tan27

=cot9-cot27+tan9-tan27

=(sec^2(9)/tan(9))-(sec^2(27)/tan(27))

=(1/(sin(9)cos(9))-((1/(sin(27)cos(27))

=(2/sin(18))-(2/(sin(54))

=4