Without using trigonometric tables, evaluate the following:
sec 17°cosec 73°+tan 68°cot 22°\dfrac{\text{sec 17°}}{\text{\text{cosec 73°}}} + \dfrac{\text{tan 68°}}{\text{cot 22°}}cosec 73°sec 17°+cot 22°tan 68° + cos2 44° + cos2 46°.
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Solving,
⇒sec 17°cosec (90° - 17°)+tan (90° - 22°)cot 22°+cos244°+cos2(90°−44°)⇒sec 17°sec 17°+cot 22°cot 22°+cos244°+sin244°⇒1+1+1⇒3.\Rightarrow \dfrac{\text{sec 17°}}{\text{cosec (90° - 17°)}} + \dfrac{\text{tan (90° - 22°)}}{\text{cot 22°}} + \text{cos}^2 44° + \text{cos}^2 (90° - 44°) \\[1em] \Rightarrow \dfrac{\text{sec 17°}}{\text{sec 17°}} + \dfrac{\text{cot 22°}}{\text{cot 22°}} + \text{cos}^2 44° + \text{sin}^2 44° \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3.⇒cosec (90° - 17°)sec 17°+cot 22°tan (90° - 22°)+cos244°+cos2(90°−44°)⇒sec 17°sec 17°+cot 22°cot 22°+cos244°+sin244°⇒1+1+1⇒3.
Hence, sec 17°cosec 73°+tan 68°cot 22°\dfrac{\text{sec 17°}}{\text{\text{cosec 73°}}} + \dfrac{\text{tan 68°}}{\text{cot 22°}}cosec 73°sec 17°+cot 22°tan 68° + cos2 44° + cos2 46° = 3.
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cosec 35° - sec 55°
cos2 26° + cos 64° sin 26° + tan 36°cot 54°\dfrac{\text{tan 36°}}{\text{cot 54°}}cot 54°tan 36°
cos 65°sin 25°+cos 32°sin 58°−sin 28° sec 62° + cosec230°\dfrac{\text{cos 65°}}{\text{sin 25°}} + \dfrac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62° + cosec}^2 30°sin 25°cos 65°+sin 58°cos 32°−sin 28° sec 62° + cosec230°
sec 29°cosec 61°+2 cot 8° cot 17° cot 45° cot 73° cot 82°−3 (sin238°+sin252°)\dfrac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 52°)cosec 61°sec 29°+2 cot 8° cot 17° cot 45° cot 73° cot 82°−3 (sin238°+sin252°).
