The value of sin2 22o-sin2 68o-2 30o is

The correct option is B 4 Now, (sin^2 22^o+sin^2 68^o+^2 30^o) =(sin^2 22^o+cos^2 22^o+^2 30^o) [ Since sin 68^o=cos 22^o ] =1+^2 30^o ...

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Standard X

Mathematics

Trigonometric Identities

The value of ...

Question

The value of $({sin}^{2} 22^{o} + {sin}^{2} 68^{o} + {cot}^{2} 30^{o})$ is

2

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4

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\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is B $4$
Now,
$({sin}^{2} 22^{o} + {sin}^{2} 68^{o} + {cot}^{2} 30^{o})$
$= ({sin}^{2} 22^{o} + {cos}^{2} 22^{o} + {cot}^{2} 30^{o})$ [ Since $sin 68^{o} = cos 22^{o}$ ]
$= 1 + {cot}^{2} 30^{o}$
$= c o s e c^{2} 30^{o}$
$= (2)^{2}$
$= 4$ .

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The value of (sin222o+sin268o+cot230o) is

The correct option is B 4Now,(sin222o+sin268o+cot230o)=(sin222o+cos222o+cot230o) [ Since sin68o=cos22o]=1+cot230o=cosec230o=(2)2=4.

The value of $({sin}^{2} 22^{o} + {sin}^{2} 68^{o} + {cot}^{2} 30^{o})$ is

The correct option is B $4$
Now,
$({sin}^{2} 22^{o} + {sin}^{2} 68^{o} + {cot}^{2} 30^{o})$
$= ({sin}^{2} 22^{o} + {cos}^{2} 22^{o} + {cot}^{2} 30^{o})$ [ Since $sin 68^{o} = cos 22^{o}$ ]
$= 1 + {cot}^{2} 30^{o}$
$= c o s e c^{2} 30^{o}$
$= (2)^{2}$
$= 4$ .