1.

If `f(theta)=(1-sin2theta+cos2theta)/(2cos2theta)`, then value of `8f(11^0)*f(34^0)`is ____

Answer» `f(theta) = (1-sin2theta+cos2theta)/(2cos2theta)`
`=((cos^2theta+sin^2theta - 2sinthetacostheta) +(cos^2theta-sin^2theta) )/(2(cos^2theta - sin^2theta))`
`= ((costheta - sintheta)^2+(costheta+sintheta)(costheta - sintheta))/(2(costheta+sintheta)(costheta - sintheta))`
`=(costheta-sintheta+costheta+sintheta)/(2(costheta+sintheta))`
`=(2costheta)/(2(costheta+sintheta))`
`=1/(1+tantheta)`
`:. f(theta) = 1/(1+tantheta)`
`:. 8f(11^@)f(34^@) = 8(1/(1+tan11^@))(1/(1+tan34^@))->(1)`
Now, `tan 34^@ = tan(45^@-11^@) = (tan45^@-tan11^@)/(1+tan45^@tan11^@) = (1-tan11^@)/(1+tan11^@)`
`:. 1+tan34^@ = 1+ (1-tan11^@)/(1+tan11^@) = 2/(1+tan11^@)`
Putting value of `1+tan34^@` in (1),
`8f(11^@)f(34^@) = 8(1/(1+tan11^@))*((1+tan11^@)/2)`
`=>8f(11^@)f(34^@) = 4.`


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