Question

$\frac{\cos 70°}{\sin 20°}+\frac{\cos 55°\mathrm{co}sec35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}=2$

Answer 1

To show:
$$\begin{gathered} \frac{\cos 70°}{\sin 20°}+\frac{\cos 55°\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°}=2 \\ \mathrm{Now},\mathrm{LHS}=\frac{\cos 70°}{\sin 20°}+\frac{\cos 55°\mathrm{cosec}35°}{\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°} \\ =\frac{\cos (90-20)°}{\sin 20°}+\frac{\cos (90-35)°\mathrm{cosec}35°}{\tan (90-85)°\tan (90-65)°\tan 45°\tan 65°\tan 85°} \\ =\frac{\sin 20°}{\sin 20°}+\frac{\sin 35°\mathrm{cosec}35°}{\cot 85\cot 65°\tan 45°\tan 65°\tan 85°}\left[\because \sin \left(90-\mathrm{\theta }\right)=\cos \mathrm{\theta }\&\tan \left(90-\mathrm{\theta }\right)=\cot \mathrm{\theta }\right] \\ =1+\frac{1}{\cot 85°\cot 65°\cot 65°\tan 65°\tan 45°} \\ =1+\frac{1}{1\times 1\times \tan 45°} \\ =1+\frac{1}{1} \\ =2 \\ =\mathrm{RHS} \end{gathered}$$