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प्रश्न
Prove the following:
`cos((3pi)/2 + x) cos(2pi + x)[cot((3pi)/2 - x) + cot(2pi + x)]` = 1
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उत्तर
L.H.S. = `cos((3pi)/2 + x) cos(2pi + x)[cot((3pi)/2 - x) + cot(2pi + x)]`
= sin x · cos x [tan x + cot x]
= `sinx * cosx[sinx/cosx + cosx/sinx]`
= `sinx*cosx[(sin^2x + cos^2x)/(sinx*cosx)]`
= `sinx*cosx xx 1/(sinx*cosx)`
= 1
= R.H.S.
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