Question

cosec (75° + θ) −sec (15° − θ) = ?
(a) 2 sec θ
(b) 2 cosec θ
(c) 0
(d) 1

Answer 1

$$\begin{gathered} (c)0.. \\ \text{We have:} \\ [\cos \text{ec}({75}^0+\theta )-\sec ({15}^0-\theta )] \\ =[\cos \text{ec}\{{90}^0-({15}^0-\theta )\}-\sec ({15}^0-\theta )] \\ \text{=sec}({15}^0-\theta )-\sec ({15}^0-\theta )[\because \cos \text{ec}({90}^0-\theta )=\sec \theta ] \\ =0 \end{gathered}$$