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Ex 9.5
Ex 9.5, 1 (ii) Important
Ex 9.5, 1 (iii)
Ex 9.5, 1 (iv)
Ex 9.5, 1 (v) Important
Ex 9.5, 1 (vi) Important
Ex 9.5, 1 (vii)
Ex 9.5, 1 (viii)
Ex 9.5, 1 (ix) Important
Ex 9.5, 1 (x)
Ex 9.5, 2 (i)
Ex 9.5, 2 (ii)
Ex 9.5, 2 (iii)
Ex 9.5, 2 (iv) Important
Ex 9.5, 2 (v)
Ex 9.5, 2 (vi) Important
Ex 9.5, 2 (vii) Important
Ex 9.5, 3 (i)
Ex 9.5, 3 (ii) Important
Ex 9.5, 3 (iii)
Ex 9.5, 3 (iv) Important
Ex 9.5, 3 (v) Important
Ex 9.5, 3 (vi)
Ex 9.5, 4 (i)
Ex 9.5, 4 (ii)
Ex 9.5, 4 (iii) Important
Ex 9.5, 4 (iv)
Ex 9.5, 4 (v) Important
Ex 9.5, 4 (vi)
Ex 9.5, 4 (vii) Important
Ex 9.5, 5 (i)
Ex 9.5, 5 (ii)
Ex 9.5, 5 (iii) Important
Ex 9.5, 5 (iv) You are here
Ex 9.5, 5 (v) Important
Ex 9.5, 6 (i)
Ex 9.5, 6 (ii) Important
Ex 9.5, 6 (iii)
Ex 9.5, 6 (iv)
Ex 9.5, 6 (v) Important
Ex 9.5, 6 (vi)
Ex 9.5, 6 (vii) Important
Ex 9.5, 6 (viii)
Ex 9.5, 6 (ix) Important
Ex 9.5, 7 (i)
Ex 9.5, 7 (ii) Important
Ex 9.5, 7 (iii)
Ex 9.5, 7 (iv) Important
Ex 9.5, 8 (i)
Ex 9.5, 8 (ii)
Ex 9.5, 8 (iii) Important
Ex 9.5, 8 (iv) Important
Last updated at March 22, 2023 by Teachoo
Ex 9.5, 5 Show that. (iv) (4ππ+3π)^2β(4ππβ3π)^2=48ππ^2 Solving (πππ+ππ)^π (π+π)^2=π^2+π^2+2ππ Putting π = 4ππ & π = 3π (π+π)^2=π^2+π^2+2ππ Putting π = 4ππ & π = 3π = (4ππ)^2+(3π)^2+2(4ππ)(3π) = 16π^2 π^2+9π^2+24ππ^2 Solving (πππβππ)^π (πβπ)^2=π^2+π^2β2ππ Putting π = 4ππ & π = 3π = (4ππ)^2+(3π)^2β2(4ππ)(3π) = 16π^2 π^2+9π^2β24ππ^2 Solving LHS (4ππ+3π)^2β(4ππβ3π)^2 = (16π^2 π^2+9π^2+24ππ^2 )β(16π^2 π^2+9π^2β24ππ^2 ) = 16π^2 π^2+9π^2+24ππ^2β16π^2 π^2β9π^2+24ππ^2 = (16π^2 π^2β16π^2 π^2 )+(9π^2β9π^2 )+(24ππ^2+24ππ^2 ) = 0+0+48ππ^2 = 48ππ^2 = R.H.S Since LHS = RHS Hence proved