What is the value of ${sin}^{2} 35^{o} + {sin}^{2} 55^{o}$ ?

0

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1

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\frac{1}{2}

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2

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Solution

The correct option is A $1$
The value of ${sin}^{2} 35^{o} + {sin}^{2} (90^{o} - 35^{o})$
$= {sin}^{2} 35^{o} + {cos}^{2} 35^{o} = 1$
$[∵ {sin}^{2} θ + {cos}^{2} θ = 1]$

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Similar questions

{sin}^{2} 65^{o} + {sin}^{2} 25^{o} + {cos}^{2} 35^{o} + {cos}^{2} 55^{o}

\frac{sec θ cos e c (90^{o} - θ) - tan θ cot (90^{o} - θ) + ({sin}^{2} 35^{o} + {sin}^{2} 55^{o})}{tan 10^{o} tan 20^{o} tan 45^{o} tan 70^{o} tan 80^{o}}

\frac{sec θ . c o s e c (90^{o} - θ) - tan θ . cot (90^{o} - θ) + {sin}^{2} 55^{o} + {sin}^{2} 35^{o}}{tan 10^{o} . tan 20^{o} . tan 60^{o} . tan 70^{o} . tan 80^{o}}

(\frac{1}{{sec}^{2} α - {cos}^{2} α} + \frac{1}{csc^{2} α - {sin}^{2} α}) \times {sin}^{2} α {cos}^{2} α = \frac{1 - {sin}^{2} α {cos}^{2} α}{2 + {sin}^{2} α {cos}^{2} α}

{cot}^{2} θ - \frac{1}{{sin}^{2} θ}

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