Question

Solve $\sin (50+\theta )-\cos (40-\theta )+\tan 1\tan 10$
$\tan 20\tan 70\tan 80\tan 89=1$

Answer 1

$\sin (50+\theta )=\sin (90-(50+\theta ))$

$=\cos (40-\theta )$

$\tan \theta \tan (90-\theta )=\tan \theta .\cot \theta =1$

LHS=

$\therefore \sin (50+\theta )-\cos (40-\theta )+\tan 1\tan 10\tan 20\tan 70\tan 80\tan 89$

$=0+\mathrm{1.1.1}$

$=1$

=RHS

Hence proved