Prove that
sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.
Solution
\[\begin{array}{l}\text{ L.H.S} = \sin( {50}^\circ + \theta) - \cos( {40}^\circ- \theta) + \tan 1^\circ \tan {10}^\circ \tan {80}^\circ \tan {89}^\circ \\ \end{array}\]
\[\begin{array}{l}= \sin{ {90}^\circ - ( {40}^\circ - \theta)} - \cos( {40}^\circ - \theta) + {\tan 1^\circ \tan( {90}^\circ - 1^\circ )}{\tan {10}^\circ \tan( {90}^\circ -{10}^{} )} \\ \end{array}\]
\[\begin{array}{l}= \cos( {40}^\circ - \theta)- \cos( {40}^\circ - \theta) + (\tan 1^\circ \cot 1^\circ )(\tan {10}^\circ \cot {10}^\circ ) \\ \end{array}\]
\[\begin{array}{l}= (\frac{1}{\cot 1^\circ } \times \cot 1^\circ )(\tan {10}^\circ \times \frac{1}{\tan {10}^0}) \\ \end{array}\]
\[\begin{array}{l}= 1 \times 1 \\ \end{array}\]
=RHS
