Question

prove :- cot(pi+x). cos(-x) /sin(pi-x). cos(pi/2+x) =cot^2x​

Answer 1

To Prove :-

$\sf \dfrac{cos ( \pi +x) cos (-x)}{sin (\pi - x ) cos \left( \dfrac{\pi}{2} +x\right)} = cot^2x$

Solution :-

$\sf L.H.S = \dfrac{cos ( \pi +x) cos (-x)}{sin (\pi - x ) cos \left( \dfrac{\pi}{2} +x\right)}$

$\bullet \bf \ cos(\pi + x) = - cos x \\\\\bullet \ cos(-x) = sinx \\\\\bullet \ sin(\pi - x) = sin x \\\\\bullet \ cos \left( \dfrac{\pi}{2}+x \right) = - sinx$

 

Substitute the values :-

            $= \sf \dfrac{-cos x \times cosx } {sin x \times - sin x } \\\\= \dfrac{-cos^2 x}{-sin^2 x } \\\\= \dfrac{cos^2x }{sin^2 x}$

$\bullet \ \bf cotx = \dfrac{cosx}{sinx}$

            $= \sf cot^2 x \\\\= R.H.S$

Hence proved.