MCQ
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$ is equal to :
- A$0$
- B$1$
- ✓$-1$
- DNone of these.
✓
Answer
Correct option: C.
$-1$
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$
$=2\big[(\sin^2\theta)^3+(\cos^2\theta)^3\big]-3\big[(\sin^2\theta)^2+(\cos^2\theta)^2\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)\big]$
$=-3\big[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta\big]$
$\{\because\ \text{a}^3+\text{b}^2=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})\}$
$=2\big[1(\sin^2\theta)^2+(\cos^2\theta)^2+2\sin^2\theta\cos^2\theta-3\sin^2\theta+\cos^2\theta\big]$
$=-3\big[(1)^2-2\sin^2\theta\cos^2\theta\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)^2-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2\big[1-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2-6\sin^2\theta\cos^3\theta-3+6\sin^2\theta\cos^2\theta$
$=-1$
$=2\big[(\sin^2\theta)^3+(\cos^2\theta)^3\big]-3\big[(\sin^2\theta)^2+(\cos^2\theta)^2\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)\big]$
$=-3\big[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta\big]$
$\{\because\ \text{a}^3+\text{b}^2=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})\}$
$=2\big[1(\sin^2\theta)^2+(\cos^2\theta)^2+2\sin^2\theta\cos^2\theta-3\sin^2\theta+\cos^2\theta\big]$
$=-3\big[(1)^2-2\sin^2\theta\cos^2\theta\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)^2-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2\big[1-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2-6\sin^2\theta\cos^3\theta-3+6\sin^2\theta\cos^2\theta$
$=-1$
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