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Prove the following:$ \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 $
To do:
We have to prove that $\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}=1$.
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
$tan\ (90^{\circ}- \theta) = cot\ \theta$
$tan\ \theta \times \cot\ \theta=1$
Therefore,
$\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$
$=\sin (90^{\circ}-(40^{\circ}-\theta))-\cos (40^{\circ}-\theta)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan (90^{\circ}-20^{\circ}) \tan (90^{\circ}-10^{\circ}) \tan (90^{\circ}-1^{\circ})$
$=\cos (40^{\circ}-\theta)-\cos (40^{\circ}-\theta)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \cot 20^{\circ} \cot 10^{\circ} \cot 1^{\circ}$
$=(\tan 1^{\circ} \cot 1^{\circ})(\tan 10^{\circ} \cot 10^{\circ})(\tan 20^{\circ} \cot 20^{\circ})$
$=1\times1\times1$
$=1$
Hence proved.
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