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प्रश्न
Evaluate the following :
`((sin 27^@)/(cos 63^@))^2 - (cos 63^@/sin 27^@)^2`
उत्तर
We have to find: `((sin 27^@)/(cos 63^@))^2 - (cos 63^@/sin 27^@)^2`
Since `sin(90^@ - theta) = cos theta` and `cos(90^@ - theta) = sintheta`
`(sin 27^@/cos 63^@)^2 - (cos 63^@/sin 27^@)^2 = (sin (90^@ - 63^@)/cos 63^@)^2 - ((cos (90^@ - 27^2))/sin 27^@)^2`
`= ((cos 63^@)/(cos 63^@))^2 - ((sin 27)/sin 27^@)^2`
= 1 -1
= 0
So value of `((sin 27^@)/cos 63^@)^2 - ((cos 63^@)/(sin 27^@))^2` is 0
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