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# Show that (i) $$tan 48° tan 23° tan 42° tan 67° \ = \ 1$$ (ii) $$cos 38° cos 52° \ - \ sin 38° sin 52 ° \ = \ 0$$

(i) $$tan 48° tan 23° tan 42 ° tan 67° \$$
$$= \ tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°\$$
$$= \ cot 42° cot 67° tan 42° tan 67°$$ [ $$\because tan(90° - \theta) \ = \ cot\theta$$ ]

$$= \ \frac{1}{tan42°} × \frac{1}{tan67°} × tan42°tan67° \ = \ 1$$

(ii) $$cos 38°cos 52°– sin 38°sin 52° \$$
$$= \ cos(90° – 52°)cos 52° – sin(90°– 52°)sin 52°$$

$$= \ sin 52° cos 52° - cos 52° sin 52° \ = \ 0$$ [$$\because cos(90°-\theta) \ = \ sin\theta$$ and $$sin(90°-\theta) \ = \ cos\theta$$ ]