Question

Prove that:

(i) tan5° tan25° tan30° tan65° tan85° = $\frac{1}{\sqrt{3}}$
(ii) cot12° cot38° cot52° cot60° cot78° = $\frac{1}{\sqrt{3}}$
(iii) cos15° cos35° cosec55° cos60° cosec75° = $\frac{1}{2}$
(iv) cos1° cos2° cos3° ... cos180° = 0
(v) ${\left(\frac{\sin 49°}{\cos 41°}\right)}^2+{\left(\frac{\cos 41°}{\sin 49°}\right)}^2=2$

Answer 1

$$\begin{gathered} \begin{array}{l}\left(\text{i}\right){\text{LHS=tan5}}^0\tan {25}^0\tan {30}^0\tan {65}^0\tan {85}^0\\ \end{array} \\ \begin{array}{l}\text{=tan}({90}^0-{85}^0)\tan ({90}^0-{65}^0)\text{×}\frac{1}{\sqrt{3}}\text{×}\frac{1}{\cot {65}^0}\frac{1}{\cot {85}^0}\\ \end{array} \\ \begin{array}{l}{\text{=cot85}}^0\cot {65}^0\frac{1}{\sqrt{3}}\frac{1}{\cot {65}^0}\frac{1}{\cot {85}^0}\\ \end{array} \\ \text{=}\frac{1}{\sqrt{3}}=\text{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=\cot 12°\cot 38°\cot 52°\cot 60°\cot 78° \\ =\tan \left(90°-12°\right)\times \tan \left(90°-38°\right)\times \cot 52°\times \frac{1}{\sqrt{3}}\times \cot 78° \\ =\frac{1}{\sqrt{3}}\times \tan 78°\times \tan 52°\times \cot 52°\times \cot 78° \\ =\frac{1}{\sqrt{3}}\times \tan 78°\times \tan 52°\times \frac{1}{\tan 52°}\times \frac{1}{\tan 78°} \\ =\frac{1}{\sqrt{3}} \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \begin{array}{l}\left(\text{iii}\right){\text{LHS=cos15}}^0\cos {35}^0\cos {\text{ec55}}^0\cos {60}^0\cos {\text{ec75}}^0\\ \end{array} \\ \begin{array}{l}\text{=cos}({90}^0-{75}^0)\cos ({90}^0-{55}^0)\frac{1}{\sin {55}^0}\text{×}\frac{1}{2}\text{×}\frac{1}{\sin {75}^0}\\ \end{array} \\ \begin{array}{l}{\text{=sin75}}^0\sin {55}^0\frac{1}{\sin {55}^0}\times \frac{1}{2}\times \frac{1}{\sin {75}^0}\\ \end{array} \\ \text{=}\frac{1}{2}=\text{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{LHS}=\cos 1°\cos 2°\cos 3°...\cos 180° \\ =\cos 1°\times \cos 2°\times \cos 3°\times ...\times \cos 90°\times ...\cos 180° \\ =\cos 1°\times \cos 2°\times \cos 3°\times ...\times 0\times ...\cos 180° \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}={\left(\frac{\sin 49°}{\cos 41°}\right)}^2+{\left(\frac{\cos 41°}{\sin 49°}\right)}^2 \\ ={\left(\frac{\cos \left(90°-49°\right)}{\cos 41°}\right)}^2+{\left(\frac{\cos 41°}{\cos \left(90°-49°\right)}\right)}^2 \\ ={\left(\frac{\cos 41°}{\cos 41°}\right)}^2+{\left(\frac{\cos 41°}{\cos 41°}\right)}^2 \\ =1^2+1^2 \\ =1+1 \\ =2 \\ =\mathrm{RHS} \end{gathered}$$

Disclaimer: The RHS of (v) given in textbook is incorrect. There should be 2 instead 1. The same has been corrected in the solution here.