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Text Solution

Solution :

`sqrt3 cosec 90^@ - sec 20^@ = 4` <br>
LHS <br>
`sqrt3 cosec 90^@ - sec 20^@` <br>
`= sqrt3 1/(sin 20^@) -1/(cos 20^@) = ( sqrt3 cos 90^@ - sin 20^@)/(sin20^@* cos20^@)` <br>
`= (sqrt3 cos 20^@ - sin 20^@)/(sin 20^@ * cos 20^@)` <br>
`= (sqrt3/2cos 20^@ - 1/2 sin 20^@)/(1/2 sin 20^@ cos20^@)` <br>
`= (sin 60^@cos20^@ - cos 60^@sin20^@)/(1/2 sin 20^@cos20^@)` <br>
`= 2 xx 2 xx (sin (60^@-20^@))/(2 sin20^@cos90^@)` <br>
`= 4 xx (sin 40^@)/(sin 40^@)` <br>
`= 4 =`RHS <br>
Hence proved**Review of some terms**

**Measure of an Angle**

**Trigonometric Ratios**

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**Relation between trigonometric ratios**

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