Question

Prove that 2 bracket open Sin power 6 theta + cos power 6 theta bracket close - 3 bracket sin power 4 theta + cos power 4 theta bracket close + 1 is equal to zero

Answer 1

Hope it helps u.......

Answer 2

Answer:

Given: $2(sin^6\theta + cos^6\theta)-3(sin^4\theta+cos^4\theta)+1$

To Prove: Given Expression equal to 0

Proof,

First consider,

$sin^6\theta + cos^6\theta = (sin^2\theta)^3 + (cos^2\theta)^3$

using identity, $a^3+b^3=(a+b)(a^2+b^2-2ab)$ we get

$\implies (sin^2\theta+cos^2\theta)(sin^4\theta+cos^4\theta-sin^2\theta \times cos^2\theta)$

$\implies 1\times(sin^4\theta+cos^4\theta-sin^2\theta \times cos^2\theta)$     ( ∵ $sin^2\theta+cos^2\theta=1$)

$\implies sin^4\theta+cos^4\theta-sin^2\theta \times cos^2\theta$ ..........(1)

Now Consider,

$sin^4\theta+cos^4\theta=(sin^2\theta)^2+(cos^2\theta)^2$

using rule, $a^2+b^2=(a+b)^2-2ab$ we get

$\implies (sin^2\theta+cos^2\theta)^2-2\, sin^2\theta\times cos^2\theta$

$\implies (1)^2-2\, sin^2\theta\times cos^2\theta$

$\implies 1-2\, sin^2\theta\times cos^2\theta$ ........ (2)

Now put value of 1. in given expression, we get

$2(sin^6\theta + cos^6\theta)-3(sin^4\theta+cos^4\theta)+1$

$\implies 2(sin^4\theta+cos^4\theta-sin^2\theta \times cos^2\theta)-3(sin^4\theta+cos^4\theta)+1$

$\implies 2\,sin^4\theta+2\,cos^4\theta-2\,sin^2\theta \times cos^2\theta-3\,sin^4\theta-3\,cos^4\theta+1$

$\implies 2\,sin^4\theta-3\,sin^4\theta+2\,cos^4\theta-3\,cos^4\theta-2\,sin^2\theta \times cos^2\theta+1$

$\implies -\,sin^4\theta-\,cos^4\theta-2\,sin^2\theta \times cos^2\theta+1$

$\implies -\,(sin^4\theta+cos^4\theta)-2\,sin^2\theta \times cos^2\theta+1$

Put value of 2

$\implies -\,(1-2\, sin^2\theta\times cos^2\theta)-2\,sin^2\theta \times cos^2\theta+1$

$\implies -\,1+2 \,sin^2\theta\times cos^2\theta-2sin^2\theta \times cos^2\theta+1$

$\implies -1+1+2\,sin^2\theta\times cos^2\theta-2sin^2\theta \times cos^2\theta$

$\implies 0$

Hence proved