Question

$sin\theta +{sin}^2\theta =1$ then P.T. ${cos}^{12}\theta +{3cos}^{10}\theta +3{cos}^8\theta +{cos}^6\theta =1$

Answer 1

$\sin \theta +{\sin }^2\theta =1$

$\Rightarrow \sin \theta =1-{\sin }^2\theta$

$\Rightarrow 1-{\sin }^2\theta =\sin \theta$

$\Rightarrow {\cos }^2\theta =\sin \theta$

$\Rightarrow {\cos }^4\theta ={\sin }^2\theta$ by squaring both sides

$\Rightarrow {\cos }^4\theta =1-{\cos }^2\theta$

$\Rightarrow {\cos }^4\theta +{\cos }^2\theta =1$

$\Rightarrow {({\cos }^4\theta +{\cos }^2\theta )}^3=1^3$

$\Rightarrow {\cos }^{12}\theta +{\cos }^6\theta +3{\cos }^4\theta \times {\cos }^2\theta ({\cos }^4\theta +{\cos }^2\theta )=1$

$\Rightarrow {\cos }^{12}\theta +{\cos }^6\theta +3{\cos }^6\theta ({\cos }^4\theta +{\cos }^2\theta )=1$

$\Rightarrow {\cos }^{12}\theta +{\cos }^6\theta +3{\cos }^{10}\theta +3{\cos }^8\theta =1$

$\therefore {\cos }^{12}\theta +3{\cos }^{10}\theta +3{\cos }^8\theta +{\cos }^6\theta =1$