Evaluate:
5 sin 66°cos 24°−2 cot 85°tan 5°\dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}cos 24°5 sin 66°−tan 5°2 cot 85°
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Solving,
⇒5 sin 66°cos 24°−2 cot 85°tan 5°\Rightarrow \dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}⇒cos 24°5 sin 66°−tan 5°2 cot 85°
By formula,
sin (90° - θ) = cos θ and cot (90° - θ) = tan θ
⇒5 sin (90° - 24°)cos 24°−2 cot (90° - 5°)tan 5°⇒5 cos 24°cos 24°−2 tan 5°tan 5°⇒5−2⇒3.\Rightarrow \dfrac{\text{5 sin (90° - 24°)}}{\text{cos 24°}} - \dfrac{\text{2 cot (90° - 5°)}}{\text{tan 5°}} \\[1em] \Rightarrow \dfrac{\text{5 cos 24°}}{\text{cos 24°}} - \dfrac{\text{2 tan 5°}}{\text{tan 5°}} \\[1em] \Rightarrow 5 - 2 \\[1em] \Rightarrow 3.⇒cos 24°5 sin (90° - 24°)−tan 5°2 cot (90° - 5°)⇒cos 24°5 cos 24°−tan 5°2 tan 5°⇒5−2⇒3.
Hence, 5 sin 66°cos 24°−2 cot 85°tan 5°\dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}cos 24°5 sin 66°−tan 5°2 cot 85° = 3.
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Evaluate :
2(tan 35°cot 55°)2+(cot 55°tan 35°)2−3(sec 40°cosec 50°)2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)2(cot 55°tan 35°)2+(tan 35°cot 55°)2−3(cosec 50°sec 40°)
sec 26° sin 64° + cosec 33°sec 57°\dfrac{\text{cosec 33°}}{\text{sec 57°}}sec 57°cosec 33°
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
Prove that :
tan (55° + x) = cot (35° - x)
