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प्रश्न
Without using trigonometric tables, find the value of the expression:
`{(sin^2 22^circ + sin^2 68^circ)/(cos^2 22^circ + cos^2 68^circ) + sin^2 63^circ + cos 63^circ sin 27^circ}`
बेरीज
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उत्तर
Given: `{(sin^2 22^circ + sin^2 68^circ)/(cos^2 22^circ + cos^2 68^circ) + sin^2 63^circ + cos 63^circ sin 27^circ}`
Step-wise calculation:
1. Note sin2 68° = sin2(90° – 22°) = cos2 22°.
Hence (sin2 22° + sin2 68°) = sin2 22° + cos2 22° = 1.
2. Also cos2 68° = cos2(90° – 22°) = sin2 22°.
So (cos2 22° + cos2 68°) = cos2 22° + sin2 22° = 1.
3. Therefore the fraction = `1/1` = 1.
4. For the remaining terms,
sin 27° = sin(90° – 63°) = cos 63°
So cos 63° · sin 27° = cos2 63°.
Thus sin2 63° + cos 63° · sin 27°
= sin2 63° + cos^2 63°
= 1
⇒ 1 + 1 = 2.
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