Question

Prove that:

$\left(i\right)tan{720}^{\circ }-cos{270}^{\circ }-sin{150}^{\circ }cos{120}^{\circ }=\frac{1}{4}$

$\left(ii\right)sin{780}^{\circ }sin{480}^{\circ }+cos{120}^{\circ }sin{150}^{\circ }=\frac{1}{2}$

$\left(iii\right)sin{780}^{\circ }sin{120}^{\circ }+cos{240}^{\circ }sin{390}^{\circ }=\frac{1}{2}$

$\left(iv\right)sin{600}^{\circ }cos{390}^{\circ }+cos{480}^{\circ }sin{150}^{\circ }=-1$

$\left(v\right)tan{250}^{\circ }cot{405}^{\circ }+tan{765}^{\circ }cot{675}^{\circ }=0$

Answer 1

$\left(i\right)LHS=tan{720}^{\circ }-cos{270}^{\circ }-sin{150}^{\circ }cos{120}^{\circ }$

$=tan4\pi -cos\left(\frac{3\pi }{2}\right)-sin(\pi -\frac{\pi }{6})$

$cos(\frac{\pi }{2}+\frac{\pi }{6})[\because \pi ={180}^{\circ }]$

$=0-0-sin\frac{\pi }{6}(-sin\frac{\pi }{6})$

$[\because tann\pi =0foralln\in Zandcos\frac{3\pi }{2}=0]$

$=si{n}^2\frac{\pi }{6}$

$=(\frac{1}{2}{)}^2=\frac{1}{4}=RHS$

Hence proved.

$\left(ii\right)LHS=sin{780}^{\circ }sin{480}^{\circ }+cos{120}^{\circ }sin{150}^{\circ }$

$=sin(4\pi +\frac{\pi }{3})sin(3\pi -\frac{\pi }{3})+cos$

$(\frac{\pi }{2}+\frac{\pi }{6})sin(\pi -\frac{\pi }{6})$

$[\because \pi ={180}^{\circ }]$

$=sin\frac{\pi }{3}\times sin\frac{\pi }{3}+(-sin\frac{\pi }{6})sin\frac{\pi }{6}$

$\left[\begin{array}{c}\because sin(4\pi +\frac{\pi }{3})=sin\frac{\pi }{3}\\ andsin(3\pi +\frac{\pi }{3})=sin\frac{\pi }{3}\end{array}\right]$

$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}-\frac{1}{4}$

$=\frac{2}{4}=\frac{1}{2}=RHS$ Hence proved.

$\left(iii\right)LHS=sin{780}^{\circ }sin{120}^{\circ }+cos{240}^{\circ }sin{390}^{\circ }$

$=sin(4\pi +\frac{\pi }{3})sin(\frac{\pi }{2}+\frac{\pi }{6})+cos(\pi +\frac{\pi }{6})sin(2\pi +\frac{\pi }{6})$

$=sin\frac{\pi }{3}\times cos\frac{\pi }{6}-cos\frac{\pi }{3}\times (+sin\frac{\pi }{6})$

$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}$

$=\frac{3}{4}-\frac{1}{4}$

$=\frac{1}{2}=\frac{1}{2}=RHS$ Hence proved.

$\left(iv\right)LHS=sin{600}^{\circ }cos{390}^{\circ }+cos{480}^{\circ }sin{150}^{\circ }$

$=sin(3\pi +\frac{\pi }{3})cos(2\pi +\frac{\pi }{6})+cos(3\pi -\frac{\pi }{3})sin(\pi -\frac{\pi }{6})$

$=-sin\frac{\pi }{3}cos\frac{\pi }{6}-cos\frac{\pi }{3}-sin\frac{\pi }{6}$

$\left[\begin{array}{c}\because sin(3\pi +\frac{\pi }{3})=-sin\frac{\pi }{3}andcos\\ (3\pi -\frac{\pi }{3})=-cos\frac{\pi }{3}\end{array}\right]$

$=\frac{-\sqrt{3}}{2}\times \frac{-sqrt3}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{-3}{4}-\frac{1}{4}=\frac{-4}{4}$

=-1 =RHS Hence proved.=

$\left(v\right)LHS=tan{225}^{\circ }cot{405}^{\circ }+tan{765}^{\circ }cot{675}^{\circ }$

$=tan(\pi +\frac{\pi }{4})cot(2\pi +\frac{\pi }{4})(4\pi +\frac{\pi }{4})cot(4\pi -\frac{\pi }{4})$

$=tan\frac{\pi }{4}cot\frac{\pi }{4}+tan\frac{\pi }{4}(-cot\frac{\pi }{4})$

=1.1 +1.(-1)=0 RHS Hence proved.