Question

Prove that

cosec (67° + θ) − sec (23° − θ) = 0

Answer 1

$$\begin{array}{l}\text{L.H.S}=cosec({67}^{\circ }+\theta )-\sec ({23}^{\circ }-\theta )\end{array}$$

$$\begin{array}{l}=cosec{90}^{\circ }-({23}^{\circ }-\theta )-\sec ({23}^{\circ }-\theta )\end{array}$$

$$\begin{array}{l}=\sec ({23}^{\circ }-\theta )-\sec ({23}^{\circ }-\theta )\\ \end{array}$$

$$=0=R.H.S.$$