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Mathematics

Without using trigonometrical tables, evaluate the following :

2 cos 27°sin 63°+tan 27°cot 63°+cos 0°\dfrac{\text{2 cos 27°}}{\text{sin 63°}} + \dfrac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°}

Trigonometrical Ratios

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Answer

Solving,

⇒2 cos 27°sin 63°+tan 27°cot 63°+cos 0°⇒2 cos 27°sin (90° - 27°)+tan 27°cot (90° - 27°)+1⇒2 cos 27°cos 27°+tan 27°tan 27°+1⇒2+1+1⇒4.\Rightarrow \dfrac{\text{2 cos 27°}}{\text{sin 63°}} + \dfrac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°} \\[1em] \Rightarrow \dfrac{\text{2 cos 27°}}{\text{sin (90° - 27°)}} + \dfrac{\text{tan 27°}}{\text{cot (90° - 27°)}} + 1 \\[1em] \Rightarrow \dfrac{\text{2 cos 27°}}{\text{cos 27°}} + \dfrac{\text{tan 27°}}{\text{tan 27°}} + 1 \\[1em] \Rightarrow 2 + 1 + 1 \\[1em] \Rightarrow 4.

Hence, 2 cos 27°sin 63°+tan 27°cot 63°+cos 0°\dfrac{\text{2 cos 27°}}{\text{sin 63°}} + \dfrac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°} = 4.

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