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प्रश्न
Prove the following:
`(cos (pi + x) cos (-x))/(sin(pi - x) cos (pi/2 + x)) = cot^2 x`
उत्तर
L.H.S. = `(cos (pi + x) cos (-x))/(sin(pi - x) cos (pi/2 + x))`
Now using `sin (π - x) = sin x, cos (pi/2+x) = - sin x`
L.H.S. = `(- cosx xx cosx )/(sinx (- sinxx)`
= `cos^2x/sin^2x = (cosx/sinx)^2`
= `cot^2xx`
= R.H.S.
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