Question

Without using trigonometric tables, prove that:
(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2
(iv) tan 15° tan 60° tan 75° = $\sqrt{3}$
(v) tan48° tan23° tan42° tan67° tan 45° = 1
(vi) (sin72° + cos18°)(sin72° − cos18°) = 0
(vii) cosec 39° cos 51° + tan 21° cot 69° – sec²21° = 0

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\mathrm{i}\right)\mathrm{LHS}=\sin {53}^0\cos {37}^0+\cos {53}^0\sin {37}^0\\ \end{array} \\ \begin{array}{l}=\sin ({90}^0-{37}^0)\cos {37}^0+\cos ({90}^0-{37}^0)\sin {37}^0\\ \end{array} \\ \begin{array}{l}=\cos {37}^0\cos {37}^0+\sin {37}^0\sin {37}^0\\ \end{array} \\ \begin{array}{l}={\cos }^2{37}^0+{\sin }^2{37}^0\\ \end{array} \\ \begin{array}{l}=1\\ \end{array} \\ \begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}\left(\mathrm{ii}\right)\mathrm{LHS}=\cos {54}^0\cos {36}^0-\sin {54}^0\sin {36}^0\\ \end{array} \\ \begin{array}{l}=\cos ({90}^0-{36}^0)\cos {36}^0-\sin ({90}^0-{36}^0)\sin {36}^0\\ \end{array} \\ \begin{array}{l}=\sin {36}^0\cos {36}^0-\cos {36}^0\sin {36}^0\\ \end{array} \\ \begin{array}{l}=0\\ \end{array} \\ \begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}\left(\mathrm{iii}\right)\mathrm{LHS}=\sec {70}^0\sin {20}^0+\cos {20}^0\mathrm{cosec}{70}^0\\ \end{array} \\ \begin{array}{l}=\sec ({90}^0-{20}^0)\sin {20}^0+\cos {20}^0\mathrm{cosec}({90}^0-{20}^0)\\ \end{array} \\ \begin{array}{l}=\mathrm{cosec}{20}^0.\frac{1}{\mathrm{cosec}{20}^0}+\frac{1}{\sec {20}^0}.\sec {20}^0\\ \end{array} \\ \begin{array}{l}=1+1\\ \end{array} \\ \begin{array}{l}=2\\ \end{array} \\ =\mathrm{RHS} \end{gathered}$$

(iv) To prove: tan15° tan60° tan75° = $\sqrt{3}$

$$\begin{gathered} \tan 15°\tan 60°\tan 75° \\ =\tan 15°\tan 60°\tan \left(90°-15°\right) \\ =\tan 15°\tan 60°\cot 15°\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =\tan 15°\tan 60°\frac{1}{\tan 15°}\left(\because \cot \theta =\frac{1}{\tan \theta }\right) \\ =\tan 60° \\ =\sqrt{3}\left(\because \tan 60°=\sqrt{3}\right) \\ \mathrm{Hence},\tan 15°\tan 60°\tan 75°=\sqrt{3}. \end{gathered}$$
$$\begin{gathered} \left(v\right)\mathrm{LHS}=\tan 48°\tan 23°\tan 42°\tan 67° \\ =\cot \left(90°-48°\right)\cot \left(90°-23°\right)\tan 42°\tan 67° \\ =\cot 42°\cot 67°\tan 42°\tan 67° \\ =\frac{1}{\tan 42°}\times \frac{1}{\tan 67°}\times \tan 42°\times \tan 67° \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{LHS}=\left(\sin 72°+\cos 18°\right)\left(\sin 72°-\cos 18°\right) \\ =\left(\sin 72°+\cos 18°\right)\left[\cos \left(90°-72°\right)-\cos 18°\right] \\ =\left(\sin 72°+\cos 18°\right)\left(\cos 18°-\cos 18°\right) \\ =\left(\sin 72°+\cos 18°\right)\left(0\right) \\ =0 \\ =\mathrm{RHS} \end{gathered}$$
(vii) To prove: cosec39° cos51° + tan21° cot69° – sec²21° = 0

$$\begin{gathered} \mathrm{cosec}39°\cos 51°+\tan 21°\cot 69°-{\sec }^{2}21° \\ =\mathrm{cosec}\left(90°-51°\right)\cos 51°+\tan 21°\cot \left(90°-21°\right)-{\sec }^{2}21° \\ =\sec 51°\cos 51°+\tan 21°\tan 21°-{\sec }^{2}21°\left(\because \cot \left(90°-\theta \right)=\tan \theta \mathrm{and}\mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ =\sec 51°\cos 51°+{\tan }^{2}21°-{\sec }^{2}21° \\ =\sec 51°\cos 51°-\left({\sec }^{2}21°-{\tan }^{2}21°\right)\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sec }^2\theta -{\tan }^2\theta =1\right) \\ =\sec 51°\cos 51°-1 \\ =\frac{1}{\cos 51°}\cos 51°-1\left(\because \sec \theta =\frac{1}{\cos \theta }\right) \\ =1-1 \\ =0 \\ \mathrm{Hence},\mathrm{cosec}39°\cos 51°+\tan 21°\cot 69°-{\sec }^{2}21°=0. \end{gathered}$$