1. Find the value of $$1 - 2 sin^{2} θ + sin^{4} θ.$$





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  • By: anil on 05 May 2019 01.55 am
    Here, $$1 - 2 sin^{2} θ + sin^{4} θ.$$ = $$1^2 + (sin^2 heta)^2 - 2 imes 1 imes sin^2 heta$$ it is similar to $$(a-b)^2$$ = $$a^2 + b^2 - 2ab$$ So, $$1^2 + (sin^2 heta)^2 - 2 imes 1 imes sin^2 heta$$ = $$(sin^2 heta - 1)^2$$......(1)
    Now $$sin^2 heta + cos^2 heta$$ = 1.....(2) From equation 1 and 2 $$(sin^2 heta - 1)^2$$ = $$(sin^2 heta - sin^2 heta - cos^2 heta)^2$$
    = $$(cos^2 heta)^2$$ = $$cos^4 heta$$
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