Question

Prove that:

$$\begin{gathered} \left(\mathrm{i}\right)\frac{\sin 70°}{\cos 20°}+\frac{\mathrm{cosec}20°}{\sec 70°}-2\cos 70°\mathrm{cosec}20°=0 \\ \left(\mathrm{ii}\right)\frac{\cos 80°}{\sin 10°}+\cos 59°\mathrm{cosec}31°=2 \\ \left(\mathrm{iii}\right)\frac{2\sin 68°}{\cos 22°}-\frac{2\cot 15°}{5\tan 75°}-\frac{3\tan 45°\tan 20°\tan 40°\tan 50°\tan 70°}{5}=1 \\ \left(\mathrm{iv}\right)\frac{\sin 18°}{\cos 72°}+\sqrt{3}\left(\tan 10°\tan 30°\tan 40°\tan 50°\tan 80°\right)=2\left[\mathrm{CBSE}2008\right] \\ \left(\mathrm{v}\right)\frac{7\cos 55°}{3\sin 35°}-\frac{4\left(\cos 70°\mathrm{cosec}20°\right)}{3\left(\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°\right)}=1 \end{gathered}$$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=\frac{\sin 70°}{\cos 20°}+\frac{\mathrm{cosec}20°}{\sec 70°}-2\cos 70°\mathrm{cosec}20° \\ =\frac{\sin 70°}{\sin \left(90°-20°\right)}+\frac{\sec \left(90°-20°\right)}{\sec 70°}-2\cos 70°\sec \left(90°-20°\right) \\ =\frac{\sin 70°}{\sin 70°}+\frac{\sec 70°}{\sec 70°}-2\cos 70°\sec 70° \\ =1+1-2\times \cos 70°\times \frac{1}{\cos 70°} \\ =2-2 \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=\frac{\cos 80°}{\sin 10°}+\cos 59°\mathrm{cosec}31° \\ =\frac{\cos 80°}{\cos \left(90°-10°\right)}+\sin \left(90°-59°\right)\mathrm{cosec}31° \\ =\frac{\cos 80°}{\cos 80°}+\sin 31°\mathrm{cosec}31° \\ =1+\sin 31°\times \frac{1}{\sin 31°} \\ =1+1 \\ =2 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{LHS}=\frac{2\sin 68°}{\cos 22°}-\frac{2\cot 15°}{5\tan 75°}-\frac{3\tan 45°\tan 20°\tan 40°\tan 50°\tan 70°}{5} \\ =\frac{2\sin 68°}{\sin \left(90°-22°\right)}-\frac{2\cot 15°}{5\cot \left(90°-75°\right)}-\frac{3\times 1\times \cot \left(90°-20°\right)\times \cot \left(90°-40°\right)\times \tan 50°\times \tan 70°}{5} \\ =\frac{2\sin 68°}{\sin 68°}-\frac{2\cot 15°}{5\cot 15°}-\frac{3\cot 70°\cot 50°\tan 50°\tan 70°}{5} \\ =2-\frac{2}{5}-\frac{3\times {\displaystyle \frac{1}{\tan 70°}}\times {\displaystyle \frac{1}{\tan 50°}}\times \tan 50°\times \tan 70°}{5} \\ =2-\frac{2}{5}-\frac{3}{5} \\ =\frac{10-2-3}{5} \\ =\frac{5}{5} \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{LHS}=\frac{\sin 18°}{\cos 72°}+\sqrt{3}\left(\tan 10°\tan 30°\tan 40°\tan 50°\tan 80°\right) \\ =\frac{\sin 18°}{\sin \left(90°-72°\right)}+\sqrt{3}\left[\cot \left(90°-10°\right)\times \frac{1}{\sqrt{3}}\times \cot \left(90°-40°\right)\times \tan 50°\times \tan 80°\right] \\ =\frac{\sin 18°}{\sin 18°}+\sqrt{3}\left(\frac{\cot 80°\times \cot 50°\times \tan 50°\times \tan 80°}{\sqrt{3}}\right) \\ =1+\left(\frac{1}{\tan 80°}\times \frac{1}{\tan 50°}\times \tan 50°\times \tan 80°\right) \\ =1+1 \\ =2 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}=\frac{7\cos 55°}{3\sin 35°}-\frac{4\left(\cos 70°\mathrm{cosec}20°\right)}{3\left(\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°\right)} \\ =\frac{7\cos 55°}{3\cos \left(90°-35°\right)}-\frac{4\left[\sin \left(90°-70°\right)\mathrm{cosec}20°\right]}{3\left[\cot \left(90°-5°\right)\times \cot \left(90°-25°\right)\times 1\times \tan 65°\times \tan 85°\right]} \\ =\frac{7\cos 55°}{3\cos 55°}-\frac{4\left(\sin 20°\mathrm{cosec}20°\right)}{3\left(\cot 85°\cot 65°\tan 65°\tan 85°\right)} \\ =\frac{7}{3}-\frac{4\left(\sin 20°\times {\displaystyle \frac{1}{\sin 20°}}\right)}{3\left({\displaystyle \frac{1}{\tan 85°}\times \frac{1}{\tan 65°}\times \tan 65°\times \tan 85°}\right)} \\ =\frac{7}{3}-\frac{4}{3} \\ =\frac{3}{3} \\ =1 \\ =\mathrm{RHS} \end{gathered}$$