Question

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 60° − sec 30° = 0
(iv) cot 34° − tan 56° = 0
(v) sin²48° + sin²42° = 1
(vi) cos²72° + cos²18° = 1

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right){\text{LHS=cos81}}^0-\sin 9^0\\ \end{array} \\ \begin{array}{l}\text{=cos}({90}^0-9^0)-\sin 9^0\\ \end{array} \\ \begin{array}{l}{\text{=sin9}}^0-\sin 9^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{ii}\right){\text{LHS=tan71}}^0-\cot {19}^0\\ \end{array} \\ \begin{array}{l}\text{=tan}({90}^0-{19}^0)-\cot {19}^0\\ \end{array} \\ \begin{array}{l}{\text{=cot19}}^0-\cot {19}^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \end{gathered}$$

(iii) To Prove: cosec60° − sec30° = 0


$$\begin{gathered} \mathrm{cosec}60°-\sec 30° \\ =\mathrm{cosec}\left(90°-30°\right)-\sec 30° \\ =\sec 30°-\sec 30°\left(\because \mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ =0 \\ \mathrm{Hence},\mathrm{cosec}60°-\sec 30°=0. \end{gathered}$$

(iv) To Prove: cot34° − tan56° = 0


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

$$\begin{gathered} \begin{array}{l}\left(\text{v}\right){\text{LHS=sin}}^2{48}^0+{\sin }^2{42}^0\\ \end{array} \\ \begin{array}{l}{\text{=sin}}^2({90}^0-{42}^0)+{\sin }^2{42}^0\\ \end{array} \\ \begin{array}{l}{\text{=cos}}^2{42}^0+{\sin }^2{42}^0\\ \end{array} \\ \begin{array}{l}\text{=1}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \end{gathered}$$


(vi) To Prove: cos²72° + cos²18° = 1


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