find the value of √3 *cosec 20 - sec 20 .
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=√3/sin20°-1/cos20°
=(√3cos20°-sin20°)/(sin20°.cos20°)
=2{(√3/2)cos20°-(1/2)sin20°}/1/2(2sin20°.cos20°)
=4{sin60°.cos20°-cos60°sin20°}/(2sin20°.cos20°)
use formula
2sinA.cosA=sin2A
and ,
sinA.cosB-cosA.sinB=sin(A-B)
hence,
=4sin(60°-20°)/(sin40°)
=4sin40°/sin40°=4
=√3/sin20°-1/cos20°
=(√3cos20°-sin20°)/(sin20°.cos20°)
=2{(√3/2)cos20°-(1/2)sin20°}/1/2(2sin20°.cos20°)
=4{sin60°.cos20°-cos60°sin20°}/(2sin20°.cos20°)
use formula
2sinA.cosA=sin2A
and ,
sinA.cosB-cosA.sinB=sin(A-B)
hence,
=4sin(60°-20°)/(sin40°)
=4sin40°/sin40°=4