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प्रश्न
If sin ∝ = `1/2` prove that (3cos∝ - `4cos^2` ∝)=0
उत्तर
𝐿𝐻𝑆 = (3𝑐𝑜𝑠𝑎 − `4 cos^3` 𝑎)
= 𝑐𝑜𝑠 𝑎(3 − `4 cos^2` 𝑎)
`= sqrt( 1 − sin^2 ∝) [3 − 4(1 − sin^2 ∝)]`
`= sqrt(1-(1/2)^2) [ 3-4 (1-(1/2)^2)]`
=`sqrt(1/1-1/4 [3-4 (1/1-1/4)])`
=`sqrt(3/4 [3-4(3/4)])`
=`sqrt(3/4 [3-3])`
=`sqrt(3/4[0])`
=0
=RHS.
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