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If `cos alpha cos beta=x/2, sinalpha sinbeta=y/2`, prove that:

`sec(alpha -ibeta)+sec(alpha-ibeta)=(4x)/(x^2-y^2)`

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#### Solution

`cos alpha cos beta=x/2 and sinalpha sinbeta=y/2`................(given)

`sec(alpha -ibeta)=1/(cos(alpha-ibeta))=1/((cos alpha cos ibeta+sinalpha sinibeta)/(cos alpha cos h beta+i sinalpha sinhbeta)) 1/(x/2+(iy)/2)` = `2/(x+iy)`..................(1)

`sec(alpha -ibeta) = 2/(x+iy)`..................(2)

from (1) and (2)

`sec(alpha -ibeta)+sec(alpha-i beta-i beta)=2/(x+iy)+2/(x+iy)=(4x)/(x^2-y^2) `

Concept: Review of Complex Numbers‐Algebra of Complex Number

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