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प्रश्न
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
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उत्तर
`sin((A + B)/2) = cos"C/2`
We know that for a triangle ΔABC
`<A + <B + <C = 180^circ`
`(<B + <A)/2 = 90^circ - (<C)/2`
`sin((A+B)/2) = sin(90^circ - C/2)`
= `cos(C/2)`
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संबंधित प्रश्न
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
