Without using trigonometric tables, evaluate the following:
sin 72°cos 18°−sec 32°cosec 58°\dfrac{\text{sin 72°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}}cos 18°sin 72°−cosec 58°sec 32°
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Solving,
⇒sin 72°cos 18°−sec 32°cosec 58°⇒sin (90° - 18°)cos 18°−sec 32°cosec (90° - 32°)As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ⇒cos 18°cos 18°−sec 32°sec 32°⇒1−1⇒0.\Rightarrow \dfrac{\text{sin 72°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} \\[1em] \Rightarrow \dfrac{\text{sin (90° - 18°)}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec (90° - 32°)}} \\[1em] \text{As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cos 18°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{sec 32°}} \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0.⇒cos 18°sin 72°−cosec 58°sec 32°⇒cos 18°sin (90° - 18°)−cosec (90° - 32°)sec 32°As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ⇒cos 18°cos 18°−sec 32°sec 32°⇒1−1⇒0.
Hence, sin 72°cos 18°−sec 32°cosec 58°=0.\dfrac{\text{sin 72°}}{\text{cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} = 0.cos 18°sin 72°−cosec 58°sec 32°=0.
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