Question

If tan θ = 1 √5 , Evaluate: cosec2θ−cot2θ cosec2θ + cot2θ

Answer 1

Solution :-

→ tan θ = 1/√5

→ Perpendicular / Base = 1/√5

so,

→ Perpendicular = P = 1

→ Base = B = √5

using pythagoras theorem,

→ Hypotenuse = H = √(B² + P²)

→ H = √[(√5)² + (1)²]

→ H = √(5 + 1)

→ H = √6

then,

→ cosec θ = H/P = √6/1 = √6

→ cot θ = 1/tan θ = 1/(1/√5) = √5

therefore,

→ (cosec²θ − cot²θ) / (cosec²θ + cot²θ)

→ [(√6)² - (√5)²] / [(√6)² + (√5)²]

→ (6 - 5)/(6 + 5)

→ 1/11 (Ans.)

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