# sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

Question:

sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

Solution:

$\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$

$=\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan \left(90^{\circ}-20^{\circ}\right) \tan \left(90^{\circ}-10^{\circ}\right) \tan \left(90^{\circ}-1^{\circ}\right)$

$=\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \cot 20^{\circ} \cot 10^{\circ} \cot 1^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \frac{1}{\tan 20^{\circ}} \frac{1}{\tan 10^{\circ}} \frac{1}{\tan 1^{\circ}} \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$

$=\cos \left(90^{\circ}-\left(50^{\circ}+\theta\right)\right)-\cos \left(40^{\circ}-\theta\right)+1 \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=\cos \left(90^{\circ}-50^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right)+1$

$=\cos \left(90^{\circ}-50^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right)+1$

$=\cos \left(40^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right)+1$

$=1$

Hence, $\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}=1$