CDS 2022 · PYQ · Trigonometry · medium
What is (sec²α + tanα·tanβ - tan²α)² + (tanα - tanβ)² - sec²α·sec²β equal to?
Answer
The correct answer is A: 0. sec²α - tan²α = 1, so first bracket = 1 + tanα·tanβ. (1+tanα tanβ)² + (tanα-tanβ)² = 1+2tanα tanβ+tan²α tan²β+tan²α-2tanα tanβ+tan²β = 1+tan²α+tan²β+tan²α tan²β = (1+tan²α)(1+tan²β) = sec²α sec²β.
- A.0✓ Correct
- B.2
- C.1
- D.-1
sec²α - tan²α = 1, so first bracket = 1 + tanα·tanβ. (1+tanα tanβ)² + (tanα-tanβ)² = 1+2tanα tanβ+tan²α tan²β+tan²α-2tanα tanβ+tan²β = 1+tan²α+tan²β+tan²α tan²β = (1+tan²α)(1+tan²β) = sec²α sec²β. So expression = 0.
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