Question

Question #47e56

Answer 1

`0.`

Explanation:

`"The Reqd. Value="2tan18^@+3sec18^@-4cos18^@,`

`={(2sin18^@)/cos18^@}-{4cos18^@-3sec18^@},`

`={(2sin18^@)/cos18^@}-{4cos18^@-3/cos18^@},`

`={(2sin18^@)/cos18^@}-{(4cos^2 18^@-3)/cos18^@},`

`={(2sin18^@)/cos18^@}-{(4cos^2 18^@-3)/cos18^@}xx(cos18^@)/(cos18^@),`

`{(2sin18^@)/cos18^@}-{(4cos^3 18^@-3cos18^@)/cos^2 18^@}......(star).`

Recall that, `4cos^3theta-3costheta=cos3theta.`

`:.(star) rArr"The Reqd. Value="{(2sin18^@)/cos18^@}-cos(3xx18^@)/cos^2 18^@,`

`={(2sin18^@)/cos18^@}-cos54^@/cos^2 18^@,`

`=(2sin18^@cos18^@-cos54^@)/cos^2 18^@.`

Since, `2sinthetacostheta=sin2theta,` we have,

`:."The Reqd. Value="(sin36^@-cos54^@)/cos^2 18^@,`

But,

`sintheta=cos(90^@-theta)rArr sin36^@=cos(90^@-36^@)=cos54^@.`

`:."The Reqd. Value="(cos54^@-cos54^@)/(cos^2 18^@)=0.`

Enjoy Maths.!