Question

$\left\{\frac{\cos 70°}{\sin 20°}+\frac{\cos 55°\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\}$

Answer 1

$$\begin{gathered} \left\{\frac{\cos 70°}{\sin 20°}+\frac{\cos 55°\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\} \\ =\left\{\frac{\cos \left(90°-20°\right)}{\sin 20°}+\frac{\cos \left(90°-35°\right)\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\} \\ =\left\{\frac{\sin 20°}{\sin 20°}+\frac{\sin 35°\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\}\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =\left\{1+\frac{\sin 35°{\displaystyle \frac{1}{\sin 35°}}}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\}\left(\because \mathrm{cosec}\theta =\frac{1}{\sin \theta }\right) \\ =\left\{1+\frac{1}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\} \\ =\left\{1+\frac{1}{\tan 5°\tan 25°\tan 45°\tan \left(90°-25°\right)\tan \left(90°-5°\right)}\right\} \\ =\left\{1+\frac{1}{\tan 5°\tan 25°\tan 45°\cot 25°\cot 5°}\right\}\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =\left\{1+\frac{1}{\tan 5°\tan 25°\tan 45°{\displaystyle \frac{1}{\tan 25°}}{\displaystyle \frac{1}{\tan 5°}}}\right\}\left(\because \cot \theta =\frac{1}{\tan \theta }\right) \\ =\left\{1+\frac{1}{\tan 45°}\right\} \\ =\left\{1+\frac{1}{1}\right\}\left(\because \tan 45°=1\right) \\ =1+1 \\ =2 \\ \mathrm{Hence},\left\{\frac{\cos 70°}{\sin 20°}+\frac{\cos 55°\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}\right\}=2. \end{gathered}$$