If for some α, β ; α ≤ β, α+β=8 and sec ²(tan ⁻¹ α)+cosec²(cot ⁻¹ β)=36, then α²+β is __________

Solution

Maths sir · 4 steps

Turn both inverse-trig terms into plain algebra: they become 1+α² and 1+β².

1. Use the identities. If θ = tan⁻¹ α, then tan θ = α, so sec² θ = 1 + tan² θ = 1 + α². In the same way, if φ = cot⁻¹ β, then cosec²φ = 1 + cot² φ = 1 + β².

   (1+α²) + (1+β²) = 36 ⇒ α² + β² = 34

2. Find the product. Square α + β = 8 to get α² + β² + 2αβ = 64. So 2αβ = 64 - 34 = 30, which gives αβ = 15.
3. Pick the values. Two numbers with sum 8 and product 15 are the roots of t² - 8t + 15 = 0, namely 3 and 5. Since α ≤ β, α = 3 and β = 5.
4. So. α² + β = 9 + 5 = 14.

The common slip: 28. Ignoring α ≤ β and taking α = 5, β = 3 gives 25 + 3 = 28, not 14.

The idea to keep. Whenever a trig function acts on an inverse trig function, write the angle as θ and use sec² θ = 1+tan² θ or cosec²θ = 1+cot² θ. The inverse trig part then disappears.

Answer from the NTA final answer key. Solution by PrepWiser.