If α + β = 90°, then the value of (1 - sin2α) (1 - cos2α) × (1 + cot2β) (1 + tan2β) is∶
If α + β = 90°, then the value of (1 - sin2α) (1 - cos2α) × (1 + cot2β) (1 + tan2β) is∶
1). 1
2). -1
3). 0
4). 2
1 answers
(1 - sin2α) (1 - cos2α) × (1 + cot2β) (1 + tan2β)
As given, α + β = 90°,
∴ α = 90° - β
And we know the formulas;
(1 + tan2β) = sec2β and (1 + cot2β) = coesce2β;
⇒ [(1 - sin2(90 - β)] [(1 - cos2(90 - β)] × cosec2β × sec2β
⇒ (1 - cos2β) (1 - sin2β) × cosec2β × sec2β
$(\Rightarrow {\rm{sin}}2{\rm{\beta }} \times {\rm{cos}}2{\rm{\beta }} \times \frac{1}{{{\rm{si}}{{\rm{n}}^2}{\rm{\beta }}}} \times \frac{1}{{{\rm{co}}{{\rm{s}}^2}{\rm{\beta }}}})$
⇒ 1
∴ (1 - sin2α) (1 - cos2α) × (1 + cot2β) (1 + tan2β) = 1Other Questions
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