prove that root 3 cosec 20 -sec20=4
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√3cosec20-sec20
= √3/sin20- 1/cos20
= √3cos20 - sin20/sin20.cos2
=4[√3/2cos20- 1/2sin20/2sin20.cos20]
=4[sin60.cos20 - cos60.sin20/sin40]-----(2sinA.cosA=sin2A)
Using identity sin(A-B) = sinA.cosB-cosA.sinB
=4sin (60-40)/sin40
=4sin40/sin40
=4
hope it helps
= √3/sin20- 1/cos20
= √3cos20 - sin20/sin20.cos2
=4[√3/2cos20- 1/2sin20/2sin20.cos20]
=4[sin60.cos20 - cos60.sin20/sin40]-----(2sinA.cosA=sin2A)
Using identity sin(A-B) = sinA.cosB-cosA.sinB
=4sin (60-40)/sin40
=4sin40/sin40
=4
hope it helps
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