Question

$({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1$ is equal to

• A. $0$
• B. $1$
• C. $-2$
• D. None of these

Answer 1

The correct option is A $0$

$2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1=2\left(\right(si{n}^2\theta {)}^3+\left(co{s}^2\theta {)}^3\right)-3(si{n}^4\theta +co{s}^4\theta )+1$

$=\left[2\right(si{n}^2\theta +co{s}^2\theta \left)\right(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta co{s}^2\theta \left)\right]-3(si{n}^4\theta +co{s}^4\theta )+1$

$=2(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta co{s}^2\theta )-3si{n}^4\theta -3co{s}^4\theta +1$

$=-si{n}^4\theta -co{s}^4\theta -2si{n}^2\theta co{s}^2\theta +1$

$=-(si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta )+1$

$=-(si{n}^2\theta +co{s}^2\theta {)}^2+1$

$=-1^2+1$

$=-1+1$

$2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1=0$