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प्रश्न
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
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उत्तर
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
`sin(50^circ + θ) = cos[90^circ - (50^circ + θ)] = cos(40^circ - θ)`
`sin(50^circ + θ) - cos(40^circ - θ)`
= `cos(40^circ - θ) - cos(40^circ - θ)`
= 0
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संबंधित प्रश्न
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Activity:
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= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
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