Solve this
Question:

$\left\{\frac{\cos 65^{\circ}}{\sin 25^{\circ}}+\frac{\operatorname{cosec} 34^{\circ}}{\sec 56^{\circ}}-\frac{2 \cos 43^{\circ} \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\}$

 

Solution:

$\left\{\frac{\cos 65^{\circ}}{\sin 25^{\circ}}+\frac{\operatorname{cosec} 34^{\circ}}{\sec 56^{\circ}}-\frac{2 \cos 43^{\circ} \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\}$

$=\left\{\frac{\cos \left(90^{\circ}-25^{\circ}\right)}{\sin 25^{\circ}}+\frac{\operatorname{cosec} 34^{\circ}}{\sec 56^{\circ}}-\frac{2 \cos \left(90^{\circ}-47^{\circ}\right) \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\}$

$=\left\{\frac{\sin 25^{\circ}}{\sin 25^{\circ}}+\frac{\operatorname{cosec} 34^{\circ}}{\sec 56^{\circ}}-\frac{2 \sin 47^{\circ} \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\} \quad\left(\because \quad \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=\left\{1+\frac{\operatorname{cosec}\left(90^{\circ}-56^{\circ}\right)}{\sec 56^{\circ}}-\frac{2 \sin 47^{\circ} \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\}$

$=\left\{1+\frac{\sec 56^{\circ}}{\sec 56^{\circ}}-\frac{2 \sin 47^{\circ} \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\} \quad\left(\because \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

$=\left\{1+1-\frac{2 \sin 47^{\circ} \frac{1}{\sin 47^{\circ}}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\} \quad\left(\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right)$

$=\left\{2-\frac{2}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\}$

$=\left\{2-\frac{2}{\tan 10^{\circ} \tan 40^{\circ} \cot 40^{\circ} \cot 10^{\circ}}\right\} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=\left\{2-\frac{2}{\tan 10^{\circ} \tan 40^{\circ} \frac{1}{\tan 40^{-}} \frac{1}{\tan 10^{-}}}\right\} \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$

$=\left\{2-\frac{2}{1}\right\}$

$=2-2$

$=0$

Hence, $\left\{\frac{\cos 65^{\circ}}{\sin 25^{\circ}}+\frac{\operatorname{cosec} 34^{\circ}}{\sec 56^{\circ}}-\frac{2 \cos 43^{\circ} \operatorname{cosec} 47^{\circ}}{\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}}\right\}=0$.

 

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