if secant theta+tan theta=x then prove that (x²+1) sin theta=x²–1
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Given that: x2+6x+8
=x2+4x+2x+8, (split middle term)
=(x2+4x)+(2x+8), (group pair of terms)
=x(x+4)+2(x+4), (factor each binomials)
=(x+4)(x+2), (factor out common factor (x+4)
Hence factors of x2+6x+8 are (x+4)(x+2)
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