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Mathematics

Without using trigonometric tables, evaluate the following:

cos 75°sin 15°+sin 12°cos 78°−cos 18°sin 72°\dfrac{\text{cos 75°}}{\text{sin 15°}} + \dfrac{\text{sin 12°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin 72°}}

Trigonometrical Ratios

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Answer

Solving,

⇒cos 75°sin 15°+sin 12°cos 78°−cos 18°sin 72°⇒cos 75°sin (90° - 75°)+sin (90° - 78°)cos 78°−cos 18°sin (90° - 18°)As, sin (90 - θ) = cos θ⇒cos 75°cos 75°+cos 78°cos 78°−cos 18°cos 18°⇒1+1−1⇒1.\Rightarrow \dfrac{\text{cos 75°}}{\text{sin 15°}} + \dfrac{\text{sin 12°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin 72°}} \\[1em] \Rightarrow \dfrac{\text{cos 75°}}{\text{sin (90° - 75°)}} + \dfrac{\text{sin (90° - 78°)}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin (90° - 18°)}} \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cos 75°}}{\text{cos 75°}} + \dfrac{\text{cos 78°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{cos 18°}} \\[1em] \Rightarrow 1 + 1 - 1 \\[1em] \Rightarrow 1.

Hence, cos 75°sin 15°+sin 12°cos 78°−cos 18°sin 72°=1.\dfrac{\text{cos 75°}}{\text{sin 15°}} + \dfrac{\text{sin 12°}}{\text{cos 78°}} - \dfrac{\text{cos 18°}}{\text{sin 72°}} = 1.

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