Without using trigonometric tables, evaluate the following:
cot 40°tan 50°−12(cos 35°sin 55°)\dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big)tan 50°cot 40°−21(sin 55°cos 35°)
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Solving,
⇒cot 40°tan 50°−12(cos 35°sin 55°)⇒cot 40°tan (90° - 40°)−12(cos 35°sin (90° - 35°))As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ⇒cot 40°cot 40°−12(cos 35°cos 35°)⇒1−12⇒12.\Rightarrow \dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big) \\[1em] \Rightarrow \dfrac{\text{cot 40°}}{\text{tan (90° - 40°)}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin (90° - 35°)}}\Big) \\[1em] \text{As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ} \\[1em] \Rightarrow \dfrac{\text{cot 40°}}{\text{cot 40°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{cos 35°}}\Big) \\[1em] \Rightarrow 1 - \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{2}.⇒tan 50°cot 40°−21(sin 55°cos 35°)⇒tan (90° - 40°)cot 40°−21(sin (90° - 35°)cos 35°)As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ⇒cot 40°cot 40°−21(cos 35°cos 35°)⇒1−21⇒21.
Hence, cot 40°tan 50°−12(cos 35°sin 55°)=12\dfrac{\text{cot 40°}}{\text{tan 50°}} - \dfrac{1}{2}\Big(\dfrac{\text{cos 35°}}{\text{sin 55°}}\Big) = \dfrac{1}{2}tan 50°cot 40°−21(sin 55°cos 35°)=21.
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tan 41°cot 49°\dfrac{\text{tan 41°}}{\text{cot 49°}}cot 49°tan 41°
cosec 17°30′sec 72° 30′\dfrac{\text{cosec 17°30}'}{\text{sec 72° 30}'}sec 72° 30′cosec 17°30′.
(sin 49°cos 41°)2+(cos 41°sin 49°)2\Big(\dfrac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\dfrac{\text{cos 41°}}{\text{sin 49°}}\Big)^2(cos 41°sin 49°)2+(sin 49°cos 41°)2
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°
