if cosθ = tanθ then
sinθ = cos^2θ
sinθ = 1 - sin^2θ
sin^2θ = 1 - sinθ
so,
2sinθ +sin^2θ +sin^3θ +sin^4θ
= 2 - 2sin^2θ + 1 - sinθ + sinθ(1 - sinθ) + (1-sinθ)^2
= 3 - 2sinθ + 1 - 2sin^2θ
= 3 - 2sinθ + sinθ - 1 + sinθ
= 2
If cosθ =tan θ , then the value of 2sinθ +sin^2θ +sin^3θ +sin^4θ is (A) 2 (B)3 (C) 1 (D) 0 (E) none of the above
4 answers
sin θ / cos θ = cos θ
so
sin θ = cos^2 θ
so
first two terms
2 cos^2 + sin^2 = cos^2 + 1
last two terms
sin^2 (sin + sin^2) = sin^2 (sin + 1 - cos^2) but cos^2 = sin
so sin^2 (sin + 1 - sin) = sin^2
so the sum is
cos^2 + 1 + sin^2 = 1+1 = 2
so
sin θ = cos^2 θ
so
first two terms
2 cos^2 + sin^2 = cos^2 + 1
last two terms
sin^2 (sin + sin^2) = sin^2 (sin + 1 - cos^2) but cos^2 = sin
so sin^2 (sin + 1 - sin) = sin^2
so the sum is
cos^2 + 1 + sin^2 = 1+1 = 2
cute problem :)
THANK YOU!!!!