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Mathematics

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)

Trigonometric Identities

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Answer

Solving,

⇒2(tan 35°cot (90° - 35°))2+(cot (90° - 35°)tan 35°)2−3(sec (90°- 50°)cosec 50°)\Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{cot (90° - 35°)}}\Big)^2 + \Big(\dfrac{\text{cot (90° - 35°)}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec (90°- 50°)}}{\text{cosec 50°}}\Big)

By formula,

cot(90° - θ) = tan θ and sec(90° - θ) = cosec θ

⇒2(tan 35°tan 35°)2+(tan 35°tan 35°)2−3(cosec 50°cosec 50°)⇒2×(1)2+(1)2−3×1⇒2+1−3⇒0.\Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 + \Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{cosec 50°}}{\text{cosec 50°}}\Big) \\[1em] \Rightarrow 2 \times (1)^2 + (1)^2 - 3 \times 1 \\[1em] \Rightarrow 2 + 1 - 3 \\[1em] \Rightarrow 0.

Hence, 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) = 0.

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