Question

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.
(i) sin 72° + cosec 72°
(ii) cosec 66° + tan 66°
(iii) tan 68° + sec 68°
(iv) cot 59° + cosec 59°
(v) cos 51° + cot 49° – sec 47°
(vi) sin 67° + cos 75°

Answer 1

(i) sin72° + cosec72°

$$\begin{gathered} \sin 72°+\mathrm{cosec}72° \\ =\sin \left(90°-18°\right)+\mathrm{cosec}\left(90°-18°\right) \\ =\cos 18°+\sec 18°\left(\because \sin \left(90°-\theta \right)=\cos \theta \mathrm{and}\mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ \mathrm{Hence},\sin 72°+\mathrm{cosec}72°=\cos 18°+\sec 18°. \end{gathered}$$

(ii) cosec66° + tan66°

$$\begin{gathered} \mathrm{cosec}66°+\tan 66° \\ =\mathrm{cosec}\left(90°-24°\right)+\tan \left(90°-24°\right) \\ =\sec 24°+\cot 24°\left(\because \tan \left(90°-\theta \right)=\cot \theta \mathrm{and}\mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ \mathrm{Hence},\mathrm{cosec}66°+\tan 66°=\sec 24°+\cot 24°. \end{gathered}$$

(iii) tan68° + sec68°

$$\begin{gathered} \tan 68°+\sec 68° \\ =\tan \left(90°-22°\right)+\sec \left(90°-22°\right) \\ =\cot 22°+\mathrm{cosec}22°\left(\because \tan \left(90°-\theta \right)=\cot \theta \mathrm{and}\sec \left(90°-\theta \right)=\mathrm{cosec}\theta \right) \\ \mathrm{Hence},\tan 68°+\sec 68°=\cot 22°+\mathrm{cosec}22°. \end{gathered}$$

(iv) cot59° + cosec59°

$$\begin{gathered} \cot 59°+\mathrm{cosec}59° \\ =\cot \left(90°-31°\right)+\mathrm{cosec}\left(90°-31°\right) \\ =\tan 31°+\sec 31°\left(\because \cot \left(90°-\theta \right)=\tan \theta \mathrm{and}\mathrm{cosec}\left(90°-\theta \right)=\sec \theta \right) \\ \mathrm{Hence},\cot 59°+\mathrm{cosec}59°=\tan 31°+\sec 31°. \end{gathered}$$

(v) cos51° + cot49° – sec47°

$$\begin{gathered} \cos 51°+\cot 49°-\sec 47° \\ =\cos \left(90°-39°\right)+\cot \left(90°-41°\right)-\sec \left(90°-43°\right) \\ =\sin 39°+\tan 41°-\mathrm{cosec}43°\left(\because \cos \left(90°-\theta \right)=\sin \theta ,\cot \left(90°-\theta \right)=\tan \theta \mathrm{and}\sec \left(90°-\theta \right)=\mathrm{cosec}\theta \right) \\ \mathrm{Hence},\cos 51°+\cot 49°-\sec 47°=\sin 39°+\tan 41°-\mathrm{cosec}43°. \end{gathered}$$

(vi) sin67° + cos75°

$$\begin{gathered} \sin 67°+\cos 75° \\ =\sin \left(90°-23°\right)+\cos \left(90°-15°\right) \\ =\cos 23°+\sin 15°\left(\because \sin \left(90°-\theta \right)=\cos \theta \mathrm{and}\cos \left(90°-\theta \right)=\sin \theta \right) \\ \mathrm{Hence},\sin 67°+\cos 75°=\cos 23°+\sin 15°. \end{gathered}$$