Question

Prove that:

$\left(i\right)\frac{cos(2\pi +\theta )cosec(2\pi +\theta )tan(\pi /2+\theta )}{sec(\pi /2+\theta )cos\theta cot(\pi +\theta )}=1$

$\left(ii\right)\frac{cosec({90}^{\circ }+\theta )+cot({450}^{\circ }+\theta )}{cosec({90}^{\circ }-\theta )+tan({180}^{\circ }-\theta )}+\frac{tan({180}^{\circ }+\theta )+sec({180}^{\circ }-\theta )}{tan({360}^{\circ }+\theta )-sec(-\theta )}=2$

$\left(iii\right)\frac{sin({180}^{\circ }+\theta )cos({90}^{\circ }+\theta )tan({270}^{\circ }-\theta )cot({360}^{\circ }-\theta )}{sin({360}^{\circ }-\theta )cos({360}^{\circ }+\theta )cosec(-\theta )sin({270}^{\circ }+\theta )}\left(iv\right)1+cot\theta -sec(\frac{\pi }{2}+\theta )\}1+cot\theta +sec(\frac{\pi }{2}+\theta \left)\right\}=2cot\theta$

$\left(v\right)\frac{tan({90}^{\circ }-\theta )sec({180}^{\circ }-\theta )sin(-\theta )}{sin({180}^{\circ }+\theta )cot({360}^{\circ }-\theta )cosec({90}^{\circ }-\theta )}=1$

Answer 1

$LHS=\frac{cos(2\pi +\theta )cosec(2\pi +\theta )tan(\pi /2+\theta )}{sec(\pi /2+\theta )cos\theta cot(\pi +\theta )}$

$=\frac{cos\theta \times cosec\theta (-cot\theta )}{-cosec\theta .cos\theta cot\theta }$

$\left[\begin{array}{c}\because tan(\frac{\pi }{2}+\theta )=-cot\theta \\ andsec(\frac{\pi }{2}+\theta )=-cosec\theta \end{array}\right]$

=1 =RHS Hence proved.

$\left(ii\right)LHS=\frac{cosec({90}^{\circ }+\theta )+cot({450}^{\circ }+\theta )}{cosec({90}^{\circ }-\theta )+tan({180}^{\circ }-\theta )}+\frac{tan({180}^{\circ }+\theta )+sec({180}^{\circ }-\theta )}{tan({360}^{\circ }+\theta )-sec(-\theta )}=2$

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

$[\because cosec({90}^{\circ }+\theta )=sec\theta ,cosec({90}^{\circ }+\theta )=sec\theta ,tan({180}^{\circ }-\theta )=-tan\theta sec(-\theta )=sec\theta ]$

$=\frac{sec\theta +cot(\frac{\pi }{2}+\theta )}{sec\theta -tan\theta }+1[\because cot(2\pi +\theta )=cot\theta ]$

$=\frac{sec\theta -tan\theta }{sec\theta -tan\theta }+1[\because cot(\frac{\pi }{2}+\theta )=-tan\theta ]$

=1+1

=2 =RHS Hence proved.

$\left(iii\right)LHS=\frac{sin({180}^{\circ }+\theta )cos({90}^{\circ }+\theta )tan({270}^{\circ }-\theta )cot({360}^{\circ }-\theta )}{sin({360}^{\circ }-\theta )cos({360}^{\circ }+\theta )cosec(-\theta )sin({270}^{\circ }+\theta )}$

$=\frac{sin\theta (-sin\theta )cot\theta (-cot\theta )}{-sin\theta cos\theta (-cosec\theta )(-cos\theta )}$

$[\because tan({270}^{\circ }-\theta )=cot\theta andsin({270}^{\circ }+\theta )=-cos\theta ]$

$=\frac{-sin\theta \times sin\theta \times cos\theta \times cos\theta \times sin\theta }{-sin\theta \times cos\theta \times sin\theta \times in\theta \times cos\theta }[\because cot\theta =\frac{cos\theta }{sin\theta }andcosec\theta =\frac{1}{sin\theta }]$

=1=RHS Hence proved.

$\left(iv\right)LHS\{1+cot\theta -sec(\frac{\pi }{2}+\theta )\}\{1+cot\theta +sec(\frac{\pi }{2}+\theta )\}$

$=\{1+cot\theta -(-cosec\theta \left)\right\}\{1+cot\theta -cosec\theta \}[\because sec(\frac{\pi }{2}+\theta )=-cosec\theta ]=\left\{\right(1+cot\theta )+cosec\theta \}\left\{\right(1+cot\theta )-cosec\theta \}$

$=(1+cot\theta {)}^2-cose{c}^2\theta [\because 1+co{t}^2\theta =cose{c}^2\theta ]$

$=cose{c}^2+2cot\theta -cosec\theta$

$=2cot\theta =RHS$ Hence proved.

$\left(v\right)LHS=\frac{tan({90}^{\circ }-\theta )sec({180}^{\circ }-\theta )sin(-\theta )}{sin({180}^{\circ }+\theta )cot({360}^{\circ }-\theta )cosec({90}^{\circ }-\theta )}$

$=\frac{cot\theta \times \left(sec\theta \right)\times (-sin\theta )}{-sin\theta \times (-cot\theta )\times sec\theta }$

=1=RHS Hence proved.