`(sec 30^(@))/(“cosec”60^(@))=?`
A. `(2)/(sqrt(3))`
B. `(sqrt(3))/(2)`
C. `sqrt(3)`
D. 1
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Correct Answer – B
`sin^(2)30^(@)+4cot^(2)45^(@)-sec^(2)60^(@)=((1)/(2))^(2)+4xx(1)^(2)-2^(2)=(1)/(4)+4-4=(1)/(4).`
Correct Answer – B
`sqrt(3)tan 2theta=3rArr tan 2theta=(3)/(sqrt(3))=sqrt(3)=tan60^(@)rArr 2theta=60^(@)rArrtheta=30^(@)`.
Correct Answer – C
Given expression `=sin47^(@)cos(90^(@)-47^(@))+cos47^(@)sin(90^(@)-47^(@))`
`=(sin^(2)47^(@)+cos^(2)47^(@))=1.`
Correct Answer – C
`2cos3theta=1rArr cos3theta=(1)/(2)=cos60^(@)rArr3theta=60^(@)rArr theta=20^(@).`
Correct Answer – D