Question

Prove that

sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.

Answer 1

$$\begin{array}{l}\text{ L.H.S}=\sin ({50}^{\circ }+\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^{\circ }\tan {10}^{\circ }\tan {80}^{\circ }\tan {89}^{\circ }\end{array}$$

$$\begin{array}{l}=\sin {90}^{\circ }-({40}^{\circ }-\theta )-\cos ({40}^{\circ }-\theta )+\tan 1^{\circ }\tan ({90}^{\circ }-1^{\circ })\tan {10}^{\circ }\tan ({90}^{\circ }-{10}^{})\end{array}$$

$$\begin{array}{l}=\cos ({40}^{\circ }-\theta )-\cos ({40}^{\circ }-\theta )+(\tan 1^{\circ }\cot 1^{\circ })(\tan {10}^{\circ }\cot {10}^{\circ })\end{array}$$
$$\begin{array}{l}=(\frac{1}{\cot 1^{\circ }}\times \cot 1^{\circ })(\tan {10}^{\circ }\times \frac{1}{\tan {10}^0})\end{array}$$

$$\begin{array}{l}=1\times 1\end{array}$$

=RHS