Question

Without using trigonometric tables, prove that:
(i) sin²12° + sin² 78° = 1
(ii) sec²29° – cot²61° = 1
(iii) tan²56° – cot²34° = 0
(iv) cos²57° – sin²33° = 0
(v) sec²50° – cot²40° = 1
(vi) cosec²72° – tan²18° = 1

Answer 1

(i) To prove: sin²12° + sin² 78° = 1

$$\begin{gathered} {\sin }^{2}12°+{\sin }^{2}78° \\ ={\left(\sin \left(90°-78°\right)\right)}^2+{\sin }^{2}78° \\ ={\cos }^{2}78°+{\sin }^{2}78°\left(\because \sin \left(90°-\theta \right)=\cos \theta \right) \\ =1\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\cos }^2\theta +{\sin }^2\theta =1\right) \\ \mathrm{Hence},{\sin }^{2}12°+{\sin }^{2}78°=1. \end{gathered}$$

(ii) To prove: sec²29° – cot²61° = 1

$$\begin{gathered} {\sec }^{2}29°-{\cot }^{2}61° \\ ={\sec }^{2}29°-{\left(\cot \left(90°-29°\right)\right)}^2 \\ ={\sec }^{2}29°-{\tan }^{2}29°\left(\because \cot \left(90°-\theta \right)=\tan \theta \right) \\ =1\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sec }^2\theta -{\tan }^2\theta =1\right) \\ \mathrm{Hence},{\sec }^{2}29°-{\cot }^{2}61°=1. \end{gathered}$$


(iii) To prove: tan²56° – cot²34° = 0

$$\begin{gathered} {\tan }^{2}56°-{\cot }^{2}34° \\ ={\left(\tan \left(90°-34°\right)\right)}^2-{\cot }^{2}34° \\ ={\cot }^{2}34°-{\cot }^{2}34°\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =0 \\ \mathrm{Hence},{\tan }^{2}56°-{\cot }^{2}34°=0. \end{gathered}$$


(iv) To prove: cos²57° – sin²33° = 0

$$\begin{gathered} {\cos }^{2}57°-{\sin }^{2}33° \\ ={\left(\cos \left(90°-33°\right)\right)}^2-{\sin }^{2}33° \\ ={\sin }^{2}33°-{\sin }^{2}33°\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =0 \\ \mathrm{Hence},{\cos }^{2}57°-{\sin }^{2}33°=0. \end{gathered}$$


(v) To prove: sec²50° – cot²40° = 1

$$\begin{gathered} {\sec }^{2}50°-{\cot }^{2}40° \\ ={\sec }^{2}50°-{\left(\cot \left(90°-50°\right)\right)}^2 \\ ={\sec }^{2}50°-{\tan }^{2}50°\left(\because \cot \left(90°-\theta \right)=\tan \theta \right) \\ =1\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sec }^2\theta -{\tan }^2\theta =1\right) \\ \mathrm{Hence},{\sec }^{2}50°-{\cot }^{2}40°=1. \end{gathered}$$


(vi) To prove: cosec²72° – tan²18° = 1

$$\begin{gathered} {\mathrm{cosec}}^{2}72°-{\tan }^{2}18° \\ ={\left(\mathrm{cosec}\left(90°-18°\right)\right)}^2-{\tan }^{2}18° \\ ={\sec }^{2}18°-{\tan }^{2}18°\left(\because \mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ =1\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sec }^2\theta -{\tan }^2\theta =1\right) \\ \mathrm{Hence},{\mathrm{cosec}}^{2}72°-{\tan }^{2}18°=1. \end{gathered}$$