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Mathematics

Without using trigonometric tables, evaluate :

2(tan 35°cot 55°)2+(cot 55°tan 35°)−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) - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)

Trigonometrical Ratios

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Answer

Solving,

2(tan 35°cot 55°)2+(cot 55°tan 35°)−3(sec 40°cosec 50°)⇒2(tan 35°cot (90° - 35°))2+(cot (90° - 35°)tan 35°)−3(sec (90° - 50°)cosec 50°)As, cot (90 - θ) = tan θ and sec (90 - θ) = cosec θ⇒2(tan 35°tan 35°)2+(tan 35°tan 35°)−3(cosec 50°cosec 50°)⇒2(1)2+1−3⇒3−3⇒0.2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)\\[1em] \Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{cot (90° - 35°)}}\Big)^2 + \Big(\dfrac{\text{cot (90° - 35°)}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec (90° - 50°)}}{\text{cosec 50°}}\Big)\\[1em] \text{As, cot (90 - θ) = tan θ and sec (90 - θ) = cosec θ} \\[1em] \Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 + \Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{cosec 50°}}{\text{cosec 50°}}\Big)\\[1em] \Rightarrow 2(1)^2 + 1 - 3 \\[1em] \Rightarrow 3 - 3 \\[1em] \Rightarrow 0.

Hence, 2(tan 35°cot 55°)2+(cot 55°tan 35°)−3(sec 40°cosec 50°)=02\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big) = 0

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