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) sin35° sin55° − cos35° cos55° = 0
(v) (sin72° + cos18°)(sin72° − cos18°) = 0
(vi) tan48° tan23° tan42° tan67° = 1

Answer 1

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

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{LHS}=\sin 35°\sin 55°-\cos 35°\cos 55° \\ =\sin 35°\cos \left(90°-55°\right)-\cos 35°\sin \left(90-55°\right) \\ =\sin 35°\cos 35°-\cos 35°\sin 35° \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}=\left(\sin 72°+\cos 18°\right)\left(\sin 72°-\cos 18°\right) \\ =\left(\sin 72°+\cos 18°\right)\left[\cos \left(90°-72°\right)-\cos 18°\right] \\ =\left(\sin 72°+\cos 18°\right)\left(\cos 18°-\cos 18°\right) \\ =\left(\sin 72°+\cos 18°\right)\left(0\right) \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{LHS}=\tan 48°\tan 23°\tan 42°\tan 67° \\ =\cot \left(90°-48°\right)\cot \left(90°-23°\right)\tan 42°\tan 67° \\ =\cot 42°\cot 67°\tan 42°\tan 67° \\ =\frac{1}{\tan 42°}\times \frac{1}{\tan 67°}\times \tan 42°\times \tan 67° \\ =1 \\ =\mathrm{RHS} \end{gathered}$$