Question

Express $\sin {81}^o+\tan {71}^o$ in terms of trigonometric ratios of angles between $0^o$ and ${45}^o$.

Answer 1

$\sin 8i+\tan 7i$

$\sin (90-9)+\tan (90-19)$

As $\sin (90-x)=\cos x$

$\tan (90-x)=\cot x$

So we get

$\cos {90}^o+\cot {19}^o$